Ask question

Ask question

All categories

Tcecarenko [31]

3 years ago

7

Which statement describes the graph?

1 answer:

Arada [10]3 years ago

8 0

Answer:
Third option:

Explanation:
The third option states that...

The graph crosses the y axis at (0,-5), which is correct. (0,-5) is the y-intercept of this graph.

The graph decreased from x=-10 to x=-2, which is also true. We can see a decrease starting from when x=-10 that ends when x=-2.

The graph remained constant from x=-2 to x=-10, which is correct. We do not see an increase or decrease on this graph from when x=-2 to when x=10.

I hope this helps! Please comment if you have any questions.

You might be interested in

The senior classes at high school A and high school B planned separate trips to New York City. The senior class at high school A

alexdok [17]

13 bus seats on each bus and 12 seats on each van

8 0

3 years ago

Of the 80 boys polled at lunch, 55% say they play video games daily. According to the poll, how many boys play video games daily

alexdok [17]

$55% = \frac{55}{100} = \frac{11}{20}$

Multiply $\frac{11}{20}$ by 80 and you will get how many boys play video games daily.

$\frac{11}{20}*80=\frac{11}{1}*4=11*4=44$

44 is the number of boys who play video games daily.

8 0

3 years ago

When you add the same number to each side of an equation, do you always get an equivalent equation? webasign?

CaHeK987 [17]

Yes you always do.

eg x - 2 = 8

x - 2 + 2 = 8 + 2

x = 10
if we plug x = 10 into the original equation we se that this fits:-

10 - 2 = 8

8 0

3 years ago

Solve for x 1/3(3x + 9) =60

juin [17]

Congrats on your first question!

Answer: x=57

///////////////////////////////////////////////////////////////////////////////////////////////////////

1/3(3x+9)=60

x+3=60

x=57

4 0

3 years ago

Read 2 more answers

Find the area under the curve y = 27/x3 from x = 1 to x = t. Evaluate the area under this curve for t = 10, t = 100, and t = 100

Darina [25.2K]

Answer:

Step-by-step explanation:

given is a function as

$y=\frac{27}{x^3}$

We are to find the area form x=1 to x=t

The curve from x=1 lies in the I quadratnt.

So area above x axis is to be calculated

Area = $\int\limits^t_1 {\frac{27}{x^3} } \, dx \\=\frac{-27}{2x^2} \\= \frac{-27}{2t^2}-\frac{-27}{2}\\=\frac{27}{2}(1-\frac{1}{t^2} )$

a) When t =10,

area = $\frac{27}{2} (1-\frac{1}{10^2} )\\= 13.365$

b) t=100

area = $\frac{27}{2} (1-\frac{1}{100^2} )\\= 13.49865$

c) t=10000

area = $\frac{27}{2} (1-\frac{1}{10000^2} )\\= 13.5$

6 0

3 years ago