All categories

3 years ago

Wich of the following is the correct graph of solution to the inequality -8 >-5x+2>-38

1 answer:

3 0

-5x + 2 > -38
-5x > -40
x < 8

-5x + 2 < -8
-5x < -10
x > 2

so the inequality becomes
2 < x < 8

so the graph will be a straight line with open circles on 2 and 8 and shading in between these values.

You might be interested in

On a map, 1 inch equals miles. Two houses are inches apart on the map. What is the actual distance between the houses? Use pen

dangina [55]

Answer:

The actual distance between the houses is 18.9 miles.

Step-by-step explanation:

The exact question is as follows :

Given - On a map, 1 inch equals 5.4 miles. Two houses are 3.5 inches apart on the map.

To find - What is the actual distance between the houses?

Solution -

Given that,

1 inch = 5.4 miles

⇒3.5 inches = 3.5 × 5.4 miles

= 18.9 miles

∴ we get

The actual distance between the houses is 18.9 miles.

7 0

2 years ago

Find the measure of z.

liraira [26]

I think it would be 120

8 0

2 years ago

Read 2 more answers

The square root of 49 is how much less than four squared

Paul [167]

The answer is 9
sqrt(49) = 7
4^2 = 16
16-7 = 9

5 0

3 years ago

Read 2 more answers

How do you do it? What do you take away and do after?

Nuetrik [128]

hi sorry I can't help with that

5 0

3 years ago

Read 2 more answers

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

Leni [432]

Answer: $\dfrac{2x^2-1}{x(x^2-1)}$

Step-by-step explanation:

The given function : $y=\ln(x(x^2 - 1)^{\frac{1}{2}})$

$\Rightarrow\ y=\ln x+\ln (x^2-1)^{\frac{1}{2}}$ [ $\because \ln(ab)=\ln a +\ln b$ ]

$\Rightarrow y=\ln x+\dfrac{1}{2}\ln (x^2-1)}$ [ $\because \ln(a)^n=n\ln a$ ]

Now , Differentiate both sides with respect to x , we will get

$\dfrac{dy}{dx}=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})\dfrac{d}{dx}(x^2-1)$ (By Chain rule)

[Note : $\dfrac{d}{dx}(\ln x)=\dfrac{1}{x}$ ]

$\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x-0)$

[ $\because \dfrac{d}{dx}(x^n)=nx^{n-1}$ ]

$=\dfrac{1}{x}+\dfrac{1}{2}(\dfrac{1}{x^2-1})(2x) = \dfrac{1}{x}+\dfrac{x}{x^2-1}\\\\\\=\dfrac{(x^2-1)+(x^2)}{x(x^2-1)}\\\\\\=\dfrac{2x^2-1}{x(x^2-1)}$

Hence, the derivative of the given function is $\dfrac{2x^2-1}{x(x^2-1)}$ .

8 0

3 years ago