An angle measures 42.2° less than the measure of its complementary angle. What is the measure of each angle?

Answer:

x=0..............................

Answer:

68.5

Step-by-step explanation:

Arc ABE = 360 - 2x

Arc AE = 2x

half the difference between the two arcs is 43

43 = [tex]\frac{1}{2}[/tex] [360 -2x - (2x)]    

Answer:

equivalent

Step-by-step explanation:

ask a question. will try to help and make it

[tex]easy \: as \: \pi[/tex]

Answer:

-3/4

Step-by-step explanation:

The line passes through the two points which are A(1,1) and B(5,-2) . We know that the slope of the line passing through two points is ,

[tex]\implies Slope =\dfrac{y_2-y_1}{x_2-x_1}\\\\\implies Slope =\dfrac{ -2-1}{5-1}\\\\\implies Slope =\dfrac{-3}{4} \\\\\implies \underline{\underline{ Slope (m) =\dfrac{-3}{4}}}[/tex]

Hence the slope of the line is -3/4 .

Answer:

7

Step-by-step explanation:

evaluate the function means substitute x=-1 and calculate

f(x) = 3x²-4x

f(-1) =f(x=-1) = 3(-1)²-4(-1) =3+4 =7, because (-1)²=1 and -4(-1) =4

Slope intercept form of a line is, y = mx + c where m is the slope and c is constant.

Judging the given equation y = 2x + 5

Slope (m) of the line is 2,

y-intercept of the line,

y = 2x + 5

y = 2×0 + 5

y = 5

Answered by GAUTHMATH

Answer:

m = 2

y intercept = 5

Step-by-step explanation:

The given equation of the line is ,

[tex]\implies y = 2x +5[/tex]

We know that the Standard equation of Slope Intercept Form of the line is,

[tex]\implies y = mx + c[/tex]

Where ,

On comparing to the Standard form of the line we get ,

[tex]\implies Slope = 2 [/tex]

[tex]\implies y - intercept= 5[/tex]

Answer:

0 and 2

Step-by-step explanation:

Answer:

use average seep hours

top = 77/12 = 6.4 (most)

middle = 6 (middle)

bottom = 50/9 = 5.5 (least)

Step-by-step explanation:

Answer:

(2x²+4x-24) in.

Step-by-step explanation:

.

Answer:

275,128

Step-by-step explanation:

"___ more than 3455" is the same as saying "___ added to 3455"

So it will look something like this:

___ + 3,455 = 278,583

Then rearrange the equation to use subtraction to solve it:

278,583 - 3,455 = ____

Plug it in the calculator or solve it by hand, and you have your answer!

(To double check, add 3,455 and 275,128, and you get 278,583)

Hope it helps (●'◡'●)

Answer:

Option b, 112°

Step-by-step explanation:

<A+<B=180

or, 68+<B=180

or, <B=112

Answered by GAUTHMATH

Answer:

(x - [tex]\sqrt{5}[/tex] )(x + [tex]\sqrt{5}[/tex] )

Step-by-step explanation:

x² - 5 ← is a difference of squares and factors in general as

a² - b² = (a - b)(a + b) , then

x² - 5

= x² - ([tex]\sqrt{5}[/tex] )²

= (x - [tex]\sqrt{5}[/tex] )(x + [tex]\sqrt{5}[/tex] )

Hi there!  

»»————-　★　————-««

I believe your answer is:  

(0.286, 5.587)

»»————-　★　————-««  

Here’s why:  

⸻⸻⸻⸻

⸻⸻⸻⸻

»»————-　★　————-««  

Hope this helps you. I apologize if it’s incorrect.  

Answer:

=x4y2z2−2x3y3z+2x3yz3+x2y4−4x2y2z2+x2z4+2xy3z−2xyz3+y2z2

Make me brainliest

Answer:

x2y-x2z-xy2+xz2+y2z-yz2

step by step

step.1

Equation at the end of step 1:(((x2)•(y-z)(+((y2)•(z-x)))+z2•(x-y)

step2

Equation at the end of step2

(((x2)•(yz))+yz•(z-x))+z2•(x-y)

step.3

equation at the end of step 3.

(x2•(y-z)+y2•(z-x))+z2•(x-y)

step4

trying to factor by pulling out:

factoring: x2y-x2z-xy2+xz2+y2z-yz2

thought fully split the expression at hand into groups,each group having two terms:

group1: y2z-yz2

group 2: x2y-x2z

group 3: xz2-yz2

pull out from each groups separately:

group 1:(x-z)•(-y2)

group 2:(y-z)•(x2)

group 3:(x-y)•(z2)

looking for common sub-expressions:

group 1:(x-z)•(-y2)

group 2:(y-z)•(x2)

group 3:( x-y)•(z2)

bad news !! factoring by pulling out fails:

The groups have no common factor and cannot be added up to form a multiplication.

final result:

x2y-x2z-xy2+xz2+y2z-yz2

1.4

2.8

3.7 by 10

*Others do by your self*

Step-by-step explanation:

1. 3×4/3= 4

2.⅖×20=8

3.6/5×7/18=7/15

4.10/7×9/5=18/7= 2 4/7. Mixed Fraction

5.4/15×25/8=5/6

6.6×¾= 9/2= 4 1/2. Mixed Fraction

Answer:

both choices with 2x+y = -3

Step-by-step explanation:

to have the solution (-4, 5), that point must be on both equations/functions, meaning it must be on either one.

in other words, if the point is not on at least one of the functions, it cannot be a solution for that system.

the given function

2x + y = -3

looks like for the point (-4, 5)

2×-4 + 5 = -3

-8 + 5 = -3

-3 = -3

correct.

but

-2x + y = -3

looks like for (-4, 5)

-2×-4 + 5 = -3

8 + 5 = -3

13 = -3

wrong. the point is not on this function.

as we can therefore rule out 2 of the answer options, the other 2 most be correct.

The two equations which when written as a system has a solution of (-4, 5) is; 2x + y = -3 and 2x + y = -3

The correct equations must have same output with the given one when we place -4 and 5 for x and y respectively.

Now, for 2x + y = -3

At x = -4, and y = 5 we have;

2(-4) + 5 = -3

Same with the right hand side.

For -2x + y = -3;

At x = -4, and y = 5 we have;

-2(-4) + 5 = 13

Not the same with the right hand side.

Thus, the two equations with 2x + y = -3 are correct

Read more about Inequalities at; https://brainly.com/question/24372553

Answer:

9. The area of rectangle S is four times bigger than rectangle R

10. (1, 3)

Step-by-step explanation:

To find the solution of the two linear equations:

y=x+2

y=-2x+5

x+2=y=-2x+5

3x=3

x=1

y=1+2

y=3

(1,3)

Answer:

-8x - 1

General Formulas and Concepts:

Algebra I

Step-by-step explanation:

Step 1: Define

Identify

-10x - 10 + 2x + 9

Step 2: Simplify

[tex]\huge\textsf{Hey there!}[/tex]

[tex]\large\textsf{-10x - 10 + 2x + 9}[/tex]

[tex]\huge\textsf{COMBINE the LIKE TERMS}[/tex]

[tex]\large\textsf{-10x + 2x - 10 + 9}[/tex]

[tex]\large\textsf{-10x + 2x}\\\\\large\textsf{ = \bf -8x}[/tex]

[tex]\large\textsf{-10 + 9}\\\\\large\textsf{ = \bf -1}[/tex]

[tex]\boxed{= \large\textsf{\bf -8x - 1}}\large\checkmark[/tex]

[tex]\boxed{\boxed{\huge\textsf{Answer: \bf -8x - 1 }}}\huge\checkmark[/tex]

[tex]\large\textsf{Good luck on your assignment and enjoy your day!}[/tex]

~[tex]\frak{Amphitrite1040:)}[/tex]

Answer:

The probability that the average score of a group of n = 4 people is between 70 and 75=0.13591

Step-by-step explanation:

We are given that

[tex]\mu=65[/tex]

[tex]\sigma=10[/tex]

n=4

We have to find the probability that the average score of a group of n = 4 people is between 70 and 75.

[tex]P(70<\bar{x}<75)=P(\frac{70-65}{\frac{10}{\sqrt{4}}}<\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{75-65}{\frac{10}{\sqrt{4}}})[/tex]

[tex]=P(\frac{5}{5}<Z<\frac{10}{5})[/tex]

[tex]=P(1<Z<2)[/tex]

[tex]=P(Z<2)-P(Z<1)[/tex]

[tex]=0.97725-0.84134[/tex]

[tex]=0.13591[/tex]

Hence,  the probability that the average score of a group of n = 4 people is between 70 and 75=0.13591

Answer:

m = 4

Step-by-step explanation:

The slope for the linear points is given by :

[tex]m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]

We have,

y₂ = 4, y₁ = 0, x₂ = 2, x₁ = 1

Putting all the values,

[tex]m=\dfrac{4-0}{2-1}\\\\\implies m=\dfrac{4}{1}\\\\m=4[/tex]

So, the slope of the line is equal to 4.

Answer:

C= 15

Step-by-step explanation:

[tex] \frac{9}{5} C + 32 = 59 [/tex]

subtracting 32 from both sides

[tex] \frac{9}{5}C = 59 -32 [/tex]

[tex] \frac{9}{5} C = 27[/tex]

Dividing both sides by 9/ 5

[tex]C = 27 \times \frac{5}{9} [/tex]

[tex]C = 3 \times 5[/tex]

[tex]C = 15[/tex]

[tex]\displaystyle\bf \underbrace{19}_910{\underbrace{1}_9} \Longrightarrow This\: is \:an \:\:arithmetic\:\: progression[/tex]

Answer:

Falso.

Step-by-step explanation:

Sea [tex]d = \frac{a}{b}[/tex] un número racional, donde [tex]a, b \in \mathbb{R}[/tex] y [tex]b \neq 0[/tex], su opuesto es un número real [tex]c = -\left(\frac{a}{b} \right)[/tex]. En el caso de elevarse a un exponente dado, hay que comprobar cinco casos:

(a) El exponente es cero.

(b) El exponente es un negativo impar.

(c) El exponente es un negativo par.

(d) El exponente es un positivo impar.

(e) El exponente es un positivo par.

(a) El exponente es cero:

Toda potencia elevada a la cero es igual a uno. En consecuencia, [tex]c = d = 1[/tex]. La proposición es verdadera.

(b) El exponente es un negativo impar:

Considérese las siguientes expresiones:

[tex]d' = d^{-n}[/tex] y [tex]c' = c^{-n}[/tex]

Al aplicar las definiciones anteriores y las operaciones del Álgebra de los números reales tenemos el siguiente desarrollo:

[tex]d' = \left(\frac{a}{b} \right)^{-n}[/tex] y [tex]c' = \left[-\left(\frac{a}{b} \right)\right]^{-n}[/tex]

[tex]d' = \left(\frac{a}{b} \right)^{(-1)\cdot n}[/tex] y [tex]c' = \left[(-1)\cdot \left(\frac{a}{b} \right)\right]^{(-1)\cdot n}[/tex]

[tex]d' = \left[\left(\frac{a}{b} \right)^{-1}\right]^{n}[/tex]y [tex]c' = \left[(-1)^{-1}\cdot \left(\frac{a}{b} \right)^{-1}\right]^{n}[/tex]

[tex]d' = \left(\frac{b}{a} \right)^{n}[/tex] y [tex]c = (-1)^{n}\cdot \left(\frac{b}{a} \right)^{n}[/tex]

[tex]d' = \left(\frac{b}{a} \right)^{n}[/tex] y [tex]c' = \left[(-1)\cdot \left(\frac{b}{a} \right)\right]^{n}[/tex]

[tex]d' = \left(\frac{b}{a} \right)^{n}[/tex] y [tex]c' = \left[-\left(\frac{b}{a} \right)\right]^{n}[/tex]

Si [tex]n[/tex] es impar, entonces:

[tex]d' = \left(\frac{b}{a} \right)^{n}[/tex] y [tex]c' = - \left(\frac{b}{a} \right)^{n}[/tex]

Puesto que [tex]d' \neq c'[/tex], la proposición es falsa.

(c) El exponente es un negativo par.

Si [tex]n[/tex] es par, entonces:

[tex]d' = \left(\frac{b}{a} \right)^{n}[/tex] y [tex]c' = \left(\frac{b}{a} \right)^{n}[/tex]

Puesto que [tex]d' = c'[/tex], la proposición es verdadera.

(d) El exponente es un positivo impar.

Considérese las siguientes expresiones:

[tex]d' = d^{n}[/tex] y [tex]c' = c^{n}[/tex]

[tex]d' = \left(\frac{a}{b}\right)^{n}[/tex] y [tex]c' = \left[-\left(\frac{a}{b} \right)\right]^{n}[/tex]

[tex]d' = \left(\frac{a}{b} \right)^{n}[/tex] y [tex]c' = \left[(-1)\cdot \left(\frac{a}{b} \right)\right]^{n}[/tex]

[tex]d' = \left(\frac{a}{b} \right)^{n}[/tex] y [tex]c' = (-1)^{n}\cdot \left(\frac{a}{b} \right)^{n}[/tex]

Si [tex]n[/tex] es impar, entonces:

[tex]d' = \left(\frac{a}{b} \right)^{n}[/tex] y [tex]c' = - \left(\frac{a}{b} \right)^{n}[/tex]

(e) El exponente es un positivo par.

Considérese las siguientes expresiones:

[tex]d' = \left(\frac{a}{b} \right)^{n}[/tex] y [tex]c' = \left(\frac{a}{b} \right)^{n}[/tex]

Si [tex]n[/tex] es par, entonces [tex]d' = c'[/tex] y la proposición es verdadera.

Por tanto, se concluye que es falso que toda potencia que se obtiene de elevar a un mismo exponente un número racional y su opuesto es la misma.

Scale is 1 inch to 6 mile.

So, ½ inch = 6 mile/2 = 3 mile

Proportional Linear Relationships can be expressed in the following ways:

From a graphical point of view, a relationship is proportional and linear if the line representing the equation goes via the origin. It is to be noted that a relationship must be linear for it to be proportional and vice versa.

Thus, it is correct to state that Proportional Linear Relationships can be expressed in the following ways:

An example of an equation that is proportional and linear is:
y = 6x + 8. Note that this linear equation is proportional because it has a constant component.

Learn more about the proportional linear relationships at;
https://brainly.com/question/2143065
#SPJ6

Answer:

were is the problem

Step-by-step explanation:

Answer:

[tex]\$3,125[/tex]

Step-by-step explanation:

Let [tex]s[/tex] represent the amount of money he initially started with.

After he spent 20% on books, he will have [tex]100\%-20\%=80\%[/tex] of his initial money left. We can represent this as [tex]0.8s[/tex].

Following that, the boy spends 20% of the remainder of his money on food. Similarly, he will have [tex]100\%-20\%=80\%[/tex] of the remainder of money he had left after he purchased the books. Therefore, he ends up with [tex]0.8(0.8s)=0.64s[/tex] of his money left.

Since we're given that he had $2,000 after all these transactions, we have the following equation:

[tex]0.64s=\$2,000[/tex]

Divide both sides by 0.64 to isolate and solve for [tex]s[/tex]:

[tex]s=\frac{2,000}{0.64}=\boxed{\$3,125}[/tex]

Therefore, the boy had $3,125 to begin with.

Answer:

2800

Step-by-step explanation:

20%+20%=40%

40/100x2000=2800

Answer:

Step-by-step explanation:

The change of base formula is

[tex]log_a(b)=\frac{log_{10}b}{log_{10}a}[/tex] so filling in our given:

[tex]log_{13}}(297)=\frac{log_{10}297}{log_{10}13}[/tex] (Note: you do not have to put the base of 10 in the change of base formula because the "normal" base for a log is 10 and that is how your calculator figures it.)

Do this on your calculator to get

[tex]\frac{log297}{log13}=2.2198[/tex] Not sure how many decimal places you needed!