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3 years ago

8

How do I find the gradient of the line y=x-5

1 answer:

Katen [24]3 years ago

8 0

The gradient is the same as the slope.
The gradient is always before the x variable or is the coefficient of the x variable when an equation is in the slope intercept form.
The gradient in this equation is 1

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Willis bought a gallon of paint.He painted a wall that is 9 feet high and 10 feet wide.Then he used he rest of the paint to pain

Otrada [13]

First you need to find the surface area of the wall that was painted, and then add the additional area in the hallway that was painted.

The wall that was painted was 9 feet by 10 feet, or 90 square feet. 90 square feet plus the additional 46 square feet in the hallway is 136 square feet, which is the total area that the paint covered.

Therefore the solution is 136 square feet.

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3 years ago

Read 2 more answers

In Jamie's drawer are 3 pairs of blue socks, 1 pair

Effectus [21]

Answer:

24%

Step-by-step explanation:

I think this is right. im sorry if its not but i went to a percent calculator and checked my answer. hope this helped :)

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3 years ago

What value should go in the empty box to complete the calculation for finding the product of 62.834 × 0.45? A)2,213,360 B)2,313,

Aleonysh [2.5K]

Answer:

28.27530

Step-by-step explanation:

Given the expression 62.834 × 0.45, to solve this expression, first we need to convert it to fraction

62.834 = 62834/1000

0.45 = 45/100

Take the product if the resulting fraction:

62.834 × 0.45 = 62834/1000 × 45/100

= (62834×45)/1000×100

= 2,827,530/100,000

= 28.27530

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3 years ago

I need help with this problem! I'll give brainlist!

aliya0001 [1]

Step-by-step explanation:

use CAH method (Cos angle = adjacent/hypotenuse)

Given adjacent = 5

Hypotenuse = 12,

$\cos(x) = \frac{adjacent}{hypotenuse} \\ \cos(x) = \frac{5}{12} \\ x = {cos}^{ - 1} ( \frac{5}{12} ) \\ = 65.4deg(nearest \: tenth)$

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3 years ago

Two vertical poles of lengths 11 ft and 14 ft stand 16 ft apart. A cable reaches from the top of one pole to some point on the g

Alex17521 [72]

We are given

total length of cable is 30 feet

and this cable is on hypotenuse sides of the triangles

so, we can draw triangle as

Let's assume those hypotenuse as 'a' and 'b'

so, we will have

$a+b=30$

now, we can find 'a' and 'b' from triangle

Smaller triangle:

$a=\sqrt{x^2+11^2}$

Larger triangle:

$b=\sqrt{(16-x)^2+14^2}$

we know that

$a+b=30$

so, we can plug this value

and we get

$\sqrt{x^2+11^2}+\sqrt{(16-x)^2+14^2} =30$

now, we can solve for x

$\left(\sqrt{x^2+11^2}\right)^2=\left(30-\sqrt{\left(16-x\right)^2+14^2}\right)^2$

$32x-1231=-60\sqrt{x^2-32x+452}$

$1024x^2-78784x+1515361=3600x^2-115200x+1627200$

now, we can use quadratic formula

we get

$x=\frac{-\sqrt{36416^2-1152389056}+36416}{5152},\:x=\frac{\sqrt{36416^2-1152389056}+36416}{5152}$

$x=4.510,x=9.627$

so,

The point should be located 4.510 feet , 9.627 feet from the smaller pole to use 30 feet cable............Answer

6 0

3 years ago