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2 years ago

6

What’s the missing length

1 answer:

mezya [45]2 years ago

7 0

Answer:

a=9in

Step-by-step explanation:

Pythagoreans Theorem

a^2+5.6^2=10.6^2

a^2+31.36=112.36

a^2=81

(take square root of both sides)

a=9

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Given the sequence -2,-8/3,-10/3,-4,-14/3 find the recursive formula

balu736 [363]

Step-by-step explanation:

-2, -8/3, -10/3, -4, -14/3

Write as multiples of 1/3.

-6/3, -8/3, -10/3, -12/3, -14/3

This is an arithmetic sequence where the first term is -6/3 and the common difference is -2/3.

Therefore, the recursive formula is:

aᵢ₊₁ = aᵢ − 2/3, a₁ = -2

3 0

3 years ago

Find the solution set of this inequality|10x+20| ≤10

Pavlova-9 [17]

Answer:

solution is

[-3,-1]

Step-by-step explanation:

we are given

$|10x+20|\leq 10$

Firstly, we will find critical values

so, let's assume it is equal

$|10x+20|= 10$

now, we can break absolute sign

For $|10x+20|= -(10x+20)$ :

$-(10x+20)= 10$

we can solve for x

$-10x-20= 10$

Add both sides by 20

$-10x-20+20= 10+20$

$-10x= 30$

Divide both sides by -10

and we get

$x=-3$

For $|10x+20|= (10x+20)$ :

$(10x+20)= 10$

we can solve for x

$10x+20= 10$

Subtract both sides by 20

$10x+20-20= 10-20$

$10x= -10$

Divide both sides by 10

and we get

$x=-1$

so, critical values are

$x=-3$

$x=-1$

now, we can draw a number line and locate these values

and then we can check inequality on each intervals

For $(-\infty,-3)$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-5

$|10\times -5+20|\leq 10$

$|-50+20|\leq 10$

$30\leq 10$

so, this is FALSE

For $[-3,-1]$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=-2

$|10\times -2+20|\leq 10$

$|-20+20|\leq 10$

$0\leq 10$

so, this is TRUE

For $(-1,\infty)$ :

We can select any random value from this interval and plug that in inequality

and we get

we can plug x=0

$|10\times 0+20|\leq 10$

$|0+20|\leq 10$

$20\leq 10$

so, this is FALSE

so, solution is

[-3,-1]

7 0

3 years ago

describe how to transform (^5 square root x^7)^3 into an expression with a rational exponent. make sure you respond with complet

diamong [38]

Answer:

$x^\frac{21}{5}$

Step-by-step explanation:

We are given an expression and we have to transform it into an expression with exponent.

5th root of x can be written as $x^\frac{1}{5}$

$(\sqrt[5]{x^7})^3\\\\((x^7)^\frac{1}{5} )^3$

The powers outside will be multiplied as: $(x^a)^b = x^{ab}$

$(x^\frac{7}{5} )^3\\x^\frac{7*3}{5}\\x^\frac{21}{5}$

where the exponent is $\frac{21}{5}$ and it is a rational number by definition of rational numbers

3 0

3 years ago

The graph below shows the hours students spent studying and their science test scores.

scoray [572]

Answer:

4.5

Step-by-step explanation:

Just took the Test

7 1

3 years ago

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Jet001 [13]

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6 0

3 years ago