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san4es73 [151]
3 years ago
12

#2 with an explanation please

Mathematics
2 answers:
Scilla [17]3 years ago
4 0
The y intercept is D) 4 because that is where the Y would be on the Y axis
Alex73 [517]3 years ago
3 0
The correct answer is D because you look at the number after the x. In this case it +4
The y int would be (0,4)

Here is another example: y=5x-2

So you would look at the number after the x which in this case is -2
The y int would be (0,-2)

Hope this helped :)
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GEOMETRY Find the perimeter of the regular polygon.<br> 3(x - 1)<br> 5x - 6
viktelen [127]

Answer:

7.5

Step-by-step explanation:

6 0
3 years ago
A is located ( 1, -1) and B is located (4, 4) Find the coordinates of point that partition
larisa86 [58]

Answer:

5 -3

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
5/9 ÷ 2/3<br> PLEASE HELP I AM ACTUALLY NOT SMART
SSSSS [86.1K]

Answer:

5/6

Step-by-step explanation:

=5/9 * 3/2

simplify

5/3 * 1/2

=

5/6

8 0
3 years ago
An engineer has designed a valve that will regulate water pressure on an automobile engine. The valve was tested on 280 engines
DENIUS [597]

Answer:

The value of the test statistic is t = -2.09

Step-by-step explanation:

The null hypothesis is:

H_{0} = 6.6

The alternate hypotesis is:

H_{1} \neq 6.6

Our test statistic is:

t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation(square roof of the variance) and n is the size of the sample.

In this problem, we have that:

X = 6.5, \mu = 6.6, \sigma = \sqrt{0.64} = 0.8, n = 280

So

t = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

t = \frac{6.5 - 6.6}{\frac{0.8}{\sqrt{280}}}

t = -2.09

The value of the test statistic is t = -2.09

4 0
3 years ago
a circular oil slick is expanding at a rate of 2m^2/h. Find the rate at which it's diameter is expanding when it's radius is 1.5
aleksandrvk [35]

Answer:  \frac{4\pi}{3} \text{ meter per hour}

Step-by-step explanation:

The circular oil slick is expanding at a rate of  2 m^2/h

Let A be the area of the circular oil slick,

So, the changes in  A with respect to time (t),

\frac{dA}{dt} = 2

\frac{d(\pi r^2)}{dt} = 2

2\pi r\frac{dr}{dt} = 2

\frac{dr}{dt} = \frac{1}{\pi r}  

Also, the change in diameter with respect to time(t),

\frac{d}{dt} (2 r) = 2 \frac{dr}{dt}

\frac{d}{dt} (2 r) = 2 \times \frac{1}{\pi r}

\frac{d}{dt} (2 r) = \frac{2}{\pi r}

For r = 1.5 m,

\frac{d}{dt} ( 2 r)]_{r=1.5} = \frac{2}{\pi \times 1.5}=\frac{20}{\pi \times 15}=\frac{4\pi }{3}\text{ meter per hour}




7 0
3 years ago
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