Find the gradient of the line segment between the points (-4,4) and (4,5).

...............................................plz help

Nezavi [6.7K]

Answer:

5 Graph. Place a for on the -2 on the Y axis. next count up 1 and right 1 and place a dot. Do the opposite going down and left. do this for however long you need.

6.

Y= -3x + 2

also not sure ;-;

7.

I think 2 and 4 but I'm not sure on this one ;-;

8.

Y = 16z (I believe)

8 days

9.

a. 2

b. 15 gallons

c. 7.5 hours

I tried my best. Hopefully this helps you :)

5 0

3 years ago

the length of the rectrangular field is twice its breadth ifvthe are of the field is 3200 square meters what is the cost of fenc

otez555 [7]

Answer:

Step-by-step explanation:

Area of a rectangle = Length x Breadth.

Area= L x B

We are told the length is twice the breadth.

You can write this as :

Length = 2 x Breadth

L = 2B

If you put that into the area formula at the beginning, we get:

Area = L x B

Area = 2B x B = $2B^2$

They also told us that the area of the field is 3200 square meters. So,

Area = $2B^2$ = 3200

Divide both sides by 2

$B^2$ = $\frac{3200}{2} = 1600$

Now to solve for B we need to square root both sides.

$B = \sqrt{1600}$ = 40

So now we know the Breadth is 40meters

Remember earlier we said Length = 2 x Breadth

$L = 80$

Cost of fencing the field = 2L + 2B

Cost of fencing the field = 2 x 80 + 2 x 40 = $240

8 0

3 years ago

Which of the following is not an expression?

miskamm [114]

$\huge \bf༆ Answer ༄$

The Correct choice among them is ~ B

$\large \boxed{3x + 4 = 7}$

because it's a equation !

4 0

3 years ago

Read 2 more answers

Find the volume and surface area of the prism below

Nata [24]

Answer:

a) The volume of the prism is 660 mm³

b) The surface area of the prism is 514 mm²

Step-by-step explanation:

The given prism for which we are to find the volume and the surface area shows the dimensions of the sides

a) To find the volume, we can consider the prism as a composite figure as follows;

The topmost cuboid with dimensions (15 - 2×3) mm, 3 mm, and 9 mm

Therefore, the volume of the topmost cuboid, V₁, is given as follows;

V₁ = 5 mm × 3 mm × 9 mm = 135 mm³

The volume of the cuboid on which the top cuboid rest, V₂, is given as follows;

V₂ = 15 mm × 5 mm × 7 mm = 525 mm³

The volume of the prism, V = V₁ + V₂

Therefore, we have;

V =135 mm³ + 525 mm³ = 660 mm³

The volume of the prism, V = 660 mm³

b) The surface area of the prism is given as follows;

The surface area of the top cuboid, SA₁, is given as follows;

SA₁ = 9 mm × 5 mm + 2 × 3 mm × 9 mm + 2 × 3 mm × 5 mm = 129 mm²

The surface area of the larger cuboid, SA₂, is given as follows;

SA₂ = 2 × 15 mm × 7 mm + 2 × 5 mm × 7 mm + 2 × 3 mm × 5 mm + 15 mm × 5 mm = 385 mm²

The surface area of the prism, SA = SA₁ + SA₂

∴ The surface area of the prism, SA = 129 mm² + 385 mm² = 514 mm²

4 0

3 years ago

The level of nitrogen oxides (NOX) in the exhaust after 50,000 miles or fewer of driving of cars of a particular model varies No

pishuonlain [190]

Answer:

The level L such that the probability that the average NOX level x for the fleet is greater than L is only 0.01 is L = 0.065.

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem:

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$