The function g is defined by the following rule.
g(x) = 5x-1

Answer:

7.2

Step-by-step explanation:

Cos θ= adjacent side÷ hypotenuse

adjacent side= 6

hypotenuse= x

θ=33°

Therefore,

cos 33°=6÷x

cos 33°= 0.8387

0.8387=6÷x

Multiply both sides by X to cancel off the X on the right hand side.

0.8387×x=6

Divide both sides by 0.8387 to subject X and to bring the values to the other side

x=6÷ 0.8387

x= 7.154

X= 7.2

Hope this helps!

Answer:

D

Step-by-step explanation:

1/2 hour -> 1/3 pool

multiply this by 2

1 hour -> 2/3 pool

so D 2/3 pool

The fraction of the pool that would be filled per hour at the given rate is: D. 2/3 of the pool.

A constant rate can be described as a quantity that changes steadlity over time.

The rate that the pool gets filled is given as:

1/2 hr = 1/3 of the water that would be filled in the pool

1 hr would be: 1/3 × 2 = 2/3

Therefore, the fraction of the pool that would be filled per hour at the given rate is: D. 2/3 of the pool.

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5/6 x 4/9 = 10/27 as a fraction form. The work for 5/6 times 4/9 as a fraction provides more insight of how to find what is 5/6 times 4/9 in fraction form and the different variations of such problems.

Answer:

y=2x+b

Step-by-step explanation:

Answer:

883.125

Step-by-step explanation:

First, use the formula for volume of a sphere (4/3 * pi * r^3) to get the volume of the full sphere, then divide it by 2 to get the volume of half of the sphere.

Hopefully this helps- let me know if you have any questions!

Answer:

Step-by-step explanation:

Given that area of an equilateral traingle 50 square units.

Let the length of side of equilateral traingle be 'x' units. To calculate length of each side we will use the formula of area of equilateral traingle:

[tex] \: \: { \underline { \boxed{\pmb { \sf{ \purple {Area_{(equilateral \: traingle)} = \dfrac{\sqrt 3}{4} \times (Side)^2}} }}}} \\ [/tex]

By substituting the values in above formula:

[tex]~[/tex]

[tex] : \implies \sf \: \: 50 = \dfrac{ \sqrt{3} }{4} \times {(x)}^{2} \\ \\ \\ : \implies \sf \: \: {(x)}^{2} = \dfrac{50 \times 4}{ \sqrt{3} } \\ \\ \\ : \implies \sf \: \: {(x)}^{2} = \dfrac{200}{ \sqrt{3} } \\ \\ \\ : \implies \sf \: \: x = \sqrt{ \dfrac{200}{ \sqrt{3} } } \\ \\ \\ : \implies \: \: { \boxed{\pmb{ \frak{ x = 10.75 \: units }}}}\\ \\ \\ [/tex]

Answer:

8/3 < 8/2

D

Step-by-step explanation:

8/3 = 2.67

8/2 = 4

So, when fractions have equivalent (same) numerators, the one with the lesser denominator is the greater fraction.

There is more left if 8 is divided by 2 than when it is divided by 3

[Answer]8/3 < 8/2

[Answer] D

PLEASE RATE!! I hope this helps!!

If you have any questions comment below

Using an exponential function, it is found that:

a) [tex]N(t) = 75(0.5)^{\frac{t}{3.8}}[/tex]

b) 37.5 grams of the gas remains after 3.8 days.

c) The amount remaining will be of 10 grams after approximately 11 days.

A decaying exponential function is modeled by:

[tex]A(t) = A(0)(1 - r)^t[/tex]

In which:

Item a:

We start with 75 grams, and then work with a half-life of 3.8 days, hence the amount after t daus is given by:

[tex]N(t) = 75(0.5)^{\frac{t}{3.8}}[/tex]

Item b:

This is N when t = 3.8, hence:

[tex]N(t) = 75(0.5)^{\frac{3.8}{3.8}} = 37.5[/tex]

37.5 grams of the gas remains after 3.8 days.

Item c:

This is t for which N(t) = 10, hence:

[tex]N(t) = 75(0.5)^{\frac{t}{3.8}}[/tex]

[tex]10 = 75(0.5)^{\frac{t}{3.8}}[/tex]

[tex](0.5)^{\frac{t}{3.8}} = \frac{10}{75}[/tex]

[tex]\log{(0.5)^{\frac{t}{3.8}}} = \log{\frac{10}{75}}[/tex]

[tex]\frac{t}{3.8}\log{0.5} = \log{\frac{10}{75}}[/tex]

[tex]t = 3.8\frac{\log{\frac{10}{75}}}{\log{0.5}}[/tex]

[tex]t \approx 11[/tex]

The amount remaining will be of 10 grams after approximately 11 days.

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Answer:

2,3

Step-by-step explanation:

you count up 2,then count over three

Answer:

Surface Area = 6×62 = 216 inches2

Step-by-step explanation:

Formula for surface area of cube is 6 into side²

EXPLANATION

Cube has all sides of equal dimensions and like that it has 6 faces.

So total surface area of cube is equal to the combined area of all the six faces.

Now each face has equal dimensions so same area

Therefore,

area of a single face =side²

Therefore, area of entire cube = 6 times the area of a single face

= 6 * side²

Answer:

yes

Step-by-step explanation:

Answer:

Step-by-step explanation:

The given information is not sufficient to determine each angle measure since the quadrilateral is irregular.

It would be good to see the picture at least.

Possible options are A or B but we can't confirm without additional detals, option C can't be a correct choice as 180° represents a straight line.

Answer:

[tex] \frac{ \alpha + \beta + \gamma }{ - d} [/tex]

Step-by-step explanation:

If we simplify that fraction, we get

[tex] \frac{ \alpha + \beta + \gamma }{ \alpha \beta \gamma } [/tex]

Keep that in mind.

If y, a ,b are zeroes of the cubic polynomial, then that means

[tex](x - \alpha )(x - \beta )(x - \gamma )[/tex]

make up the polynomial.

Notice that leading xoeffeicent will be 1, so the roots will multiply to

[tex] - d[/tex]

so

[tex] \alpha \beta \gamma = - d[/tex]

which gives us

[tex] \frac{ \alpha + \beta + \gamma }{ - d} [/tex]

Proof:

Consider the function

[tex](x - 2)(x - 3)(x - 5)[/tex]

The roots are 2, 3, 5.

D is -30 so we get

Using the value,

[tex] \frac{2 + 3 + 5}{ 30} = \frac{1}{3} [/tex]

If we use the orginal equation, we get

[tex] \frac{1}{6} + \frac{1}{10} + \frac{1}{15} = \frac{10}{30} = \frac{1}{3} [/tex]

Answer:

Hey,mate

Notice that leading xoeffeicent will be 1, so the roots will multiply to

The roots are 2, 3, 5.

[tex]\sqrt{2} \sqrt{3} \sqrt{5}[/tex]

D is -30

Answer:

3

Step-by-step explanation:

As you can see when you list out the factors of each number, 3 is the greatest number that 39 and 57 divides into.

The function f(x) = x² and g(x) = |x|² are the same since they both give a positive value.

From the complete information, it should be noted that the function f(x) = x² and g(x) = |x|² are the same since they both give a positive value.

Also, f(x) = x³ and g(x) = |x|³ are not equal. This is because both functions are not equal for x < 0. This can be illustrated when x = -1. In this case, f(x) = -1 and g(x) = 1.

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The function f(x) = x² and g(x) = |x|² are the same since they both give a positive value.

Two functions are equal if they have the same domain and codomain and their values are the same for all elements of the domain.

The function f(x) = x² and g(x) = |x|² are the same since they both give a positive value.

Also, f(x) = x³ and g(x) = |x|³ are not equal.

This is because both functions are not equal for x < 0.

This can be illustrated when x = -1. In this case, f(x) = -1 and g(x) = 1.

Hence, the function f(x) = x² and g(x) = |x|² are the same since they both give a positive value.

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#SPJ4

Answer:Y = Co + C*x.22.45 = Co + C*200,Co = 22.45 - 200C.29.65 = Co + C*350.Co = 29.65 - 350C. = 22.45 - 200C150C = 7.20, C = 0.048/min.Co = 22.45 - 200*0.048 = $12.85/mo. = Initial cost.Y = 0.048x + 12.85.2. Y = 0.048*110 + 12.85 = $18.13.3. 34.10 = 0.048x + 12.85, X = ?.

Step-by-step explanation:

The answer to your question is 69

Given that the cyclist rides distances of 5 miles and 10 miles due to the wind, the equation that can be used to calculate the speed is therefore;

[tex]\frac{10}{v + 4} = \frac{5}{v - 4} [/tex]

The distance the cyclist rides against the wind = 5 miles

The distance the cyclist rides with the wind = 10 miLes

Let t represent the time it took the cyclist in both directions, we have;

[tex]t = \frac{10}{v + 4} = \frac{5}{v - 4} [/tex]

Where;

v = The cyclist's average rate of speed

Which gives;

10 × (v - 4) = 5 × (v - 4)

10v - 40 = 5v - 20

5v = 20

v = 20 ÷ 5 = 5

The equation that can be used to calculate the cyclist's average rate of speed is therefore;

[tex]t = \frac{10}{v + 4} = \frac{5}{v - 4} [/tex]

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Answer:

C.

Step-by-step explanation:

red: 3/8 sheet per toy

20 × 3/8 = 60/8 = 56/8 + 4/8 = 7 1/2

yellow: 5/8 sheet per toy

20 × 5/8 = 100/8 = 50/4 = 25/2 = 12 1/2

Since he needs 7 1/2 and 12 1/2, he needs to get 8 and 13 becasue 7 and 12 are not enough.

Answer: C.

Answer:

7.8 feet

Step-by-step explanation:

Use the formula of tan to find his height.

tan Θ = [tex]\frac{opposite}{adjacent}[/tex]

tan 48 = [tex]\frac{x}{7}[/tex]

x = tan 48 × 7

x = 7.8 ft

The yearly salary after paying for taxes, insurances and other expenses is  r808.42.

The first step is to determine the total fraction used to pay for taxes, insurances and other expenses:

1/2 + 1/16 + 5/6

(48 + 6 + 80) / 96 = 134 / 96=  1 19/48

Now subtract 1 19/48 from 1

1 19/48 - 1 = 19/48

Finally, divide 19/48 by 320

320 x 48/19 = r808.42

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Answer: sin is negative, cos is positive

Step-by-step explanation:

Answer:

Step-by-step explanation:

sin0= 0.8

Answer: The answer should be 164,340.

Answer: 20.4% decrease

Step-by-step explanation:

First, find how many fewer cats were adopted out this year: 1,526 - 1214 = 312.

Now, find what percent of 1,526 312 is: 1,526 * x = 312 → x = .204 * 100 = 20.4% :)

Answer:

the side s is 18

Step-by-step explanation:

s=?

Area of square(A)=324

Now,

A=s²

324=s²

√(324)=s

s=18

Step-by-step explanation:

Table A represents an arithmetic sequence with a common difference d=+7.5

so for x=4, it's y³+7.5=25+7.5=32.5

for x=5, it's y⁴+7.5=32.5+7.5=40

Table B represents a geometric sequence with a common ratio r=4, so you're essentially multiplying by 4.

for x=4, y⁴=y³×4=160×4=640

for x=5, y⁵=y⁴×4=640×4=2560

➝ Hypotenuse of triangle ( a ) = 21.63 mm

➝ Hypotenuse of triangle ( b ) = 150 mm

➝ Hypotenuse of triangle ( c ) = 111.80 mm

[tex] \quad\rule{300pt}{1.5pt}\quad[/tex]

We have to find the length of hypotenuse in the given 3 triangles, which can be done by using Pythagoras theorem.

" In a right angled triangle, the square of hypotenuse side is equal to the sum of square of other two sides "

[tex] \qquad \bull \:{\pmb{\mathfrak{ h^2 = b^2 + p^2}}}[/tex]

And we have to convert the answer to the units indicated in red i.e, in mm.

Since 1cm = 10 mm, we will convert the given values of length of side in mm before putting the values in the formula

[tex] :\implies\qquad \sf{ h^2 = b^2 + p^2}[/tex]

[tex] :\implies\qquad \sf{ h =\sqrt{b^2 + p^2}}[/tex]

[tex] :\implies\qquad \sf{h=\sqrt{ (12)^2 + (18)^2 }}[/tex]

[tex] :\implies\qquad \sf{ h= \sqrt{144 + 324}}[/tex]

[tex] :\implies\qquad \sf{ h = \sqrt{468}}[/tex]

[tex] :\implies\qquad\underline{\underline{\pmb{ \sf{ h = 21.63 mm}}}}[/tex]

[tex] :\implies\qquad \sf{ h^2 = b^2 + p^2}[/tex]

[tex] :\implies\qquad \sf{h =\sqrt{b^2 + p^2} }[/tex]

[tex] :\implies\qquad \sf{ h = \sqrt{(90)^2+(120)^2}}[/tex]

[tex] :\implies\qquad \sf{ h=\sqrt{8100+14400}}[/tex]

[tex] :\implies\qquad \sf{ h =\sqrt{22500}}[/tex]

[tex] :\implies\qquad\underline{\underline{\pmb{ \sf{h = 150mm}}} }[/tex]

[tex] :\implies\qquad \sf{h^2 = b^2 + p^2 }[/tex]

[tex] :\implies\qquad \sf{ h=\sqrt{b^2 + p^2}}[/tex]

[tex] :\implies\qquad \sf{ h =\sqrt{(100)^2)+(50)^2}}[/tex]

[tex] :\implies\qquad \sf{ h=\sqrt{10000+2500}}[/tex]

[tex] :\implies\qquad \sf{ h =\sqrt{12500}}[/tex]

[tex] :\implies\qquad \underline{\underline{\pmb{\sf{h = 111.80mm}}} }[/tex]

Hypotenuse of triangle ( a ) = 21.63 mm

Hypotenuse of triangle ( b ) = 150 mm

Hypotenuse of triangle ( c ) = 111.80 mm

find the length of hypotenuse in the given 3 triangles, which can be done by using Pythagoras theorem.

[tex]h^2 = b^2 + p^2[/tex]

And we have to convert the answer to the units indicated in red i.e, in mm.

Since 1cm = 10 mm, we will convert the given values of length of side in mm before putting the values in the formula

[tex]For \: \: triangle ( a )

\qquad \sf{ h^2 = b^2 + p^2}[/tex]

[tex]\qquad \sf{ h =\sqrt{b^2 + p^2}}[/tex][tex]\qquad \sf{h=\sqrt{ (12)^2 + (18)^2 }}[/tex][tex]\qquad\sf{h=\sqrt{ (12)^2 + (18)^2 }} \\ \\ \qquad \sf{ h= \sqrt{144 + 324}} \\ \\ \qquad \sf{ h = \sqrt{468}}

\\ \\\qquad\underline{\underline{\pmb{ \sf{ h = 21.63 mm}}}} \\ \\ For \: \: triangle ( b ) \qquad \sf{ h^2 = b^2 + p^2} \\ \\ \qquad \sf{h =\sqrt{b^2 + p^2} } \\ \\ \qquad \sf{ h = \sqrt{(90)^2+(120)^2}} \\ \\ \qquad \sf{ h=\sqrt{8100+14400}} \\ \\

\qquad \sf{ h =\sqrt{22500}} \\ \\\qquad\underline{\underline{\pmb{ \sf{h = 150mm}}} } \\ \\ For \: \: triangle ( c ) \qquad \sf{h^2 = b^2 + p^2 } \\ \\ \qquad \sf{ h=\sqrt{b^2 + p^2}} \\ \\\qquad\sf{ h=\sqrt{(100)^2)+(50)^2}} \\ \\\qquad\sf{ h=\sqrt{10000+2500}} \\ \\ \qquad \sf{ h =\sqrt{12500}} \\ \\

\qquad\underline{\underline{\pmb{\sf{h = 111.80mm}}} } [/tex]

Answer:

Step-by-step explanation:

The relations describing the two numbers can be written as equations. First, we need to assign variables: let x and y represent the first and second numbers, respectively.

The given relations are then ...

  7x +6y = 31

  3x -10y = 29

__

Such a system of equations can be solved many ways. One that is usually convenient is to use a graphing calculator. It tells us ...

Answer:  Distributive property  a(b+c) = ab + ac

-5(7)x + 7(7) = -35x + 49 (D)