Anthony worked to earn $25.00. Joyce worked for $8.00 per hour. If he earns $5.00 per hour, how many hours did Anthony work?

m angle APD.

Hope It's help you.

Step-by-step explanation:

yeah angle m is APD so yeah

Answer:

B. Leg-Acute (LA)

E. Angle-Angle-Side (AAS)

Step-by-step explanation:

Congruence is the relationship between two or more shapes with respect to their common properties.

Comparing the properties of triangles LMN and OPQ, it would be observed that two angles are similar and one side.

So that the congruence theorems or postulates required are:

B. Leg-Acute (LA)

E. Angle-Angle-Side (AAS)

Answer:

B

Step-by-step explanation:

9 divided by 4, then multiply by 12

Answer:

y = 2.25(12)

Step-by-step explanation:

$9 for 4 pizzas

$9/4 = $2.25

$9 for 4 pizzas is the same ratio as $2.25 per 1 pizza

To deliver x number of pizzas, the charge, y, is

y = 2.25x

For 12 pizzas, the delivery charge is

y = 2.25(12)

Answer:

A is wrong

Step-by-step explanation:

Answer:

A(-a)*(-1)=(-a)

Step-by-step explanation:

Anything multiplied by 0 is 0. Therefore, option B is correct.

Option A is wrong because when multiplying two negatives you should get a positive not a negative.

Anything multiplied by 0 is 0. Therefore, option B is correct

Option C is correct because a negative times a positive gives a positive.

Option D is correct because you can't solve for different times being multiplied you can only combine like terms therefore your answer will equal the same result here.

Answer:

x = 4 - 5/2 y

Step-by-step explanation:

2x + 5y = 8.

Subtract 5x from each side

2x + 5y -5y = 8-5y

2x = 8 -5y

Divide each side by 2

2x/2 = (8-5y)/2

x = 4 - 5/2 y

Answer:

x=[tex]\frac{-5y}{2}[/tex]+4

Step-by-step explanation:

Hi there!

We're given the equation 2x+5y=8 and we want to solve it for x

as we don't know the value of y, x won't actually be a number (it'll be an expression instead). However, we will solve this as if it was a regular equation by isolating x onto one side

here's the given equation

2x+5y=8

start by subtracting 5y from both sides

2x=-5y+8

divide both sides by 2

x=[tex]\frac{-5y}{2}[/tex]+4

Hope this helps!

Answer:

Step-by-step explanation:

Step-by-step explanation:

Given that,

The bag of skittles has 3 parts red skittles, 5 parts yellow skittles and 7 parts orange skittle.

Ratio notation is= 3:5:7

Frction notation,

Red skittle[tex]=\dfrac{3}{3+5+7}=\dfrac{1}{5}[/tex]

Yellow skittle [tex]=\dfrac{5}{3+5+7}=\dfrac{1}{3}[/tex]

Orange skittle [tex]=\frac{7}{3+5+7}=\dfrac{7}{15}[/tex]

Hence, this is the required solution.

Answer:

Step-by-step explanation:

First use the fact that the sum of interior angles of a triangle is 180° to find the measure of the missing angle

in ΔDEF we have 180 -68 -81 = 31°

in ΔABC we have 180 -68 -33 = 79°

the measure of the angles in ΔABC and ΔDEF are different 33°, 68°, 79° ≠81°, 68°, 31°so the two triangles are not congruent.

Answer:

your answer is 1275

Step-by-step explanation:

ccifufbfjrorwfeejeidiffgfkxxocyegdo

Answer:

[tex]{ \tt{a < b→true}} \\ { \bf{reason :because \: - 1 \: is \: less \: than \: 4.5 }}\\ \\ { \tt{ |a| > b →false}} \\ { \bf{reason :1 \: is \: not \: greater \: than \: 4.5 }}\\ \\ { \tt{a < |b| →true}}[/tex]

Answer:

true

false

true

Step-by-step explanation:

Answer:

angle 1 = 105º

angle 2  = 75º

Step-by-step explanation:

angle 2  = 75º

corresponding angle to the 75º angle

angle 1 = 105º

it's supplementary to angle 2

180 - 75 = 105

Answer:

d

Step-by-step explanation:

it is impossible to tell since no numbers were given nor signs were offered to help us know whether we are dealing with additon , subtraction , multiplication or division

Answer:

y=5, y=[tex]\frac{38}{11}[/tex]

Step-by-step explanation:

Hi there!

We are given the equation

[tex]\frac{y+2}{y-3}[/tex]+[tex]\frac{y-1}{y-4}[/tex]=[tex]\frac{15}{2}[/tex] and we need to solve for y

first, we need to find the domain, which is which is the set of values that y CANNOT be, as the denominator of the fractions cannot be 0

which means that y-3≠0, or y≠3, and y-4≠0, or y≠4

[tex]\frac{y+2}{y-3}[/tex] and [tex]\frac{y-1}{y-4}[/tex] are algebraic fractions, meaning that they are fractions (notice the fraction bar), but BOTH the numerator and denominator have algebraic expressions

Nonetheless, they are still fractions, and we need to add them.

To add fractions, we need to find a common denominator

One of the easiest ways to find a common denominator is to multiply the denominators of the fractions together

Let's do that here;

on [tex]\frac{y+2}{y-3}[/tex], multiply the numerator and denominator by y-4

[tex]\frac{(y+2)(y-4)}{(y-3)(y-4)}[/tex]; simplify by multiplying the binomials together using FOIL to get:

[tex]\frac{y^{2}-2y-8}{y^{2}-7y+12}[/tex]

Now on [tex]\frac{y-1}{y-4}[/tex], multiply the numerator and denominator by y-3

[tex]\frac{(y-1)(y-3)}{(y-4)(y-3)}[/tex]; simplify by multiplying the binomials together using FOIL to get:

[tex]\frac{y^{2}-4y+3}{y^{2}-7y+12}[/tex]

now add [tex]\frac{y^{2}-2y-8}{y^{2}-7y+12}[/tex] and [tex]\frac{y^{2}-4y+3}{y^{2}-7y+12}[/tex] together

Remember: since they have the same denominator, we add the numerators together

[tex]\frac{y^{2}-2y-8+y^{2}-4y+3}{y^{2}-7y+12}[/tex]

simplify by combining like terms

the result is:

[tex]\frac{2y^{2}-6y-5}{y^{2}-7y+12}[/tex]

remember, that's set equal to [tex]\frac{15}{2}[/tex]

here is our equation now:

[tex]\frac{2y^{2}-6y-5}{y^{2}-7y+12}[/tex]=[tex]\frac{15}{2}[/tex]

it is a proportion, so you may cross multiply

2(2y²-6y-5)=15(y²-7y+12)

do the distributive property

4y²-12y-10=15y²-105y+180

subtract 4y² from both sides

-12y-10=11y²-105y+180

add 12 y to both sides

-10=11y²-93y+180

add 10 to both sides

11y²-93y+190=0

now we have a quadratic equation

Let's solve this using the quadratic formula

Recall that the quadratic formula is y=(-b±√(b²-4ac))/2a, where a, b, and c are the coefficients of the numbers in a quadratic equation

in this case,

a=11

b=-93

c=190

substitute into the formula

y=(93±√(8649-4(11*190))/2*11

simplify the part under the radical

y=(93±√289)/22

take the square root of 289

y=(93±17)/22

split into 2 separate equations:

y=[tex]\frac{93+17}{22}[/tex]

y=[tex]\frac{110}{22}[/tex]

y=5

and:

y=[tex]\frac{93-17}{22}[/tex]

y=[tex]\frac{76}{22}[/tex]

y=[tex]\frac{38}{11}[/tex]

Both numbers work in this case (remember: the domain is y≠3, y≠4)

So the answer is:

y=5, y=[tex]\frac{38}{11}[/tex]

Hope this helps! :)

Answer:

∠XWV ∠UWV

Hope this helps! :)

rounding 0.0692494 to 3SF is 0.069

Answer:

first question answer is step 2 and second step is answer step 3, last one is step is answer step 1

D. 7

Hope this helps! :)

______________

Answer:

The cube root of 343 is 7.

Answer:

it helps you balance your checkbook

Step-by-step explanation:

Balancing your checkbook means that you do a check in your account that shows how much money is available.

I hope it helps❤️

Answer:

5

Step-by-step explanation:

Greater than 0

Less than 100

Is not 4, but is greater than four

Is less than 5.5

Must be a whole number

5 is the only whole number between 4 and 5.5

Answer:

195 sq cm

Step-by-step explanation:

15 x 15 = 225

----------------------

3(8) + 2(3) = 30

----------------------

225 - 30 = 195 sq cm

Answer:

Step-by-step explanation:

The average rate of change is the slope. Slope has a formula that is the change in y over the change in x, which is a fraction. The only time a fraction can have a vlue of 0 is where the numerator of the fraction is equal to 0 (since we are not allowed to have a denominator of 0). If the change in y is in the top of the slope fraction, then we have to find the interval where the y values are the same. I'll show you one where the y values are not the same so you can compare it to the slope where the y values are the same. We will find the slope of choice A.

When x = -3, y = 0 so the coordinate is (-3, 0).

When x = 5, y = 5 so the coordinate is (5,4). Now let's find the slope (aka average rate of change) between those 2 coordinates:

[tex]m=\frac{4-0}{5-(-3)}=\frac{4}{8}=\frac{1}{2}[/tex]  and the top of the fraction is a 1, not a 0, so the average rate of change between these 2 points is 1/2, not 0. Now let's do D.

When x = -3, y = 0 so the coordinate is (-3, 0).

When x = -1, y = 0 so the coordinate is (-1, 0). The slope between these 2 points is

[tex]m=\frac{0-0}{-1-(-3)}=\frac{0}{2}=0[/tex] This fraction is equal to 0 because the numerator is 0. Choice D is the one you want.

Answer:

100 times for 1

Step-by-step explanation:

If the cube is a fair one, you would expect it to land on each number about 1/6 of the time. Since 1 is one of the faces, you would expect to land on it 1/6 of the time.

(1/6)*600=100

So any number should land 100 times

Answer:

[tex]1\frac{6}{9}[/tex]

Step-by-step explanation:

[tex]\frac{8}{9}+\frac{7}{9}[/tex]

[tex]\frac{8+7}{9}=\frac{15}{9}=1\frac{6}{9}[/tex]

Answer:

[tex]\text{1. }2.53^{\circ},\\\text{2. }792,772.98\:\mathrm{ft}[/tex]

Step-by-step explanation:

Start by converting miles to feet:

[tex]150\text{ miles}=150\cdot 5280\text{ feet}=792000\text{ feet}[/tex]

Form a right triangle with the plane's displacement marking the hypotenuse of this triangle. We can now use basic trig for a right triangle to solve for the angle of descent.

Let the angle of descent be [tex]\theta[/tex]. In a right triangle, the tangent of an angle is equal to its opposite side divide by its adjacent side.

Therefore, we have:

[tex]\tan \theta=\frac{35000}{792000},\\\\\theta =\arctan(\frac{35000}{792000})=2.53036411\approx \boxed{2.53^{\circ}}[/tex]

In all right triangles, [tex]a^2+b^2=c^2[/tex], where [tex]c[/tex] is the hypotenuse of the triangle and [tex]a[/tex] and [tex]b[/tex] are the two legs of the triangle (Pythagorean Theorem).

Therefore, the length of the plane's path is:

[tex]35,000^2+792,000^2=c^2,\\c^2=628489000000,\\c=\sqrt{628489000000}=792772.981376\approx \boxed{792,772.98\:\mathrm{ft}}[/tex]

Answer:

x(x-x[tex]y^{2}[/tex]+y)

Step-by-step explanation:

Answer:91%

Step-by-step explanation:

Answer:

B. 1 : 1

Step-by-step explanation:

the triangle has the same size of legs (the same angles 45°)

[tex]\underline \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }[/tex]

[tex]\huge\underline{\sf{\red{Problem:}}}[/tex]

[tex]\underline{\sf{\red{Formula\:for\: area\: of \:rectangular\: form:}}}[/tex]

[tex]\quad\quad\quad\quad\boxed{\sf{\red{➢} \: \sf{a=l×b}} }[/tex]

[tex]\underline{\sf{\red{Given:}}}[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{Length\:of\:rectangular\:form = 45m} }[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{Breadth\:of\:rectangular\:form = 15m} }[/tex]

[tex]\underline{\sf{\red{Solution:}}}[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{a=l×b} }[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{a=45m×15m} }[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{a=\boxed{\sf{{675m}^{2}}}} }[/tex]

[tex]\underline{\sf{\red{For\:rectangular\: form\:with\:path:}}}[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{Length\:of\:rectangular\:form = 45m+2=47m} }[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{Breadth\:of\:rectangular\:form = 15m+2=17m} }[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{a=l×b} }[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{a=47m×17m} }[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{a=\boxed{\sf{{799m}^{2}}}} }[/tex]

[tex]\huge\underline{\sf{\red{Area\:of\:path:}}}[/tex]

[tex]\quad\quad\quad\quad\sf{\red{➢} \: \sf{a={799m}^{2}-{675m}^{2}} }[/tex]

[tex]\quad\quad\quad\quad\boxed{\sf{\red{➢} \: \sf{a={124m}^{2}} }}[/tex]

[tex]\huge\underline{\sf{\red{Answer:}}}[/tex]

[tex]\huge\quad\quad\underline{\boxed{\sf{\red{a={124m}^{2}} }}}[/tex]

[tex]\underline \bold{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }[/tex]

[tex]\sf{\red{︎✰}ShineBrighter\red{︎✰}}[/tex]

[tex]\sf{✍︎ C.Rose\red{❀}}[/tex]

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