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4 years ago

7. Ned took a test with 25 questions. He lost 4

Mathematics

1 answer:

Alenkinab [10]4 years ago

5 0

84 points lost overall

You might be interested in

(5h-4)-(h-5) combining like terms

murzikaleks [220]

Distribute the negative throughout the set of parenthesis

5h-4-h+5

Ok, so your like terms are...

5h and -h

-4 and +5

If you rearrange the terms so that the like terms are next to each other, it is easier to combine them:

5h-h-4+5

5h-h=4h

-4+5=1

So, the problem is:

4h+1

Hope that helps :)

5 0

4 years ago

1.What is the slope of (0,2) and (-8,-8). What is the slope of (-8,-5) and (8,15)

ss7ja [257]

Answer:

Slope of (0,2) and (-8,-8): 5/4

Slope of (-8,-5) and (8,15): 5/4

Both slopes are 5/4

m = 5/4

Step-by-step explanation:

To find the slope:

1. Use the formula $m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ .

2. Assign point 1 and point 2

3. State (x₁ , y₁) and (x₂, y₂)

4. Substitute x₁ , y₁, x₂, and y₂ into the formula

5. Solve

Slope of (0,2) and (-8,-8):

Point 1: (0,2) x₁ = 0 y₁ = 2

Point 2: (-8,-8) x₂ = -8 y₂ = -8

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ Formula for slope

$m = \frac{-8-2}{-8-0}$ Substituted points

$m = \frac{-10}{-8}$ Simplified the subtraction then the negatives

$m = \frac{10}{8}$ Reduce fraction by dividing top and bottom by 2

$m = \frac{5}{4}$ First slope

Slope of (-8,-5) and (8,15):

Point 1: (-8,-5) x₁ = -8 y₁ = -5

Point 2: (8,15) x₂ = 8 y₂ = 15

$m = \frac{y_{2}-y_{1}}{x_{2}-x_{1}}$ Formula for slope

$m = \frac{15-(-5)}{8-(-8)}$ Substituted points

$m = \frac{15+5}{8+8}$ Simplified the subtraction into addition

$m = \frac{20}{16}$ Reduce fraction (divide top and bottom by 4)

$m = \frac{5}{4}$ Second slope

3 0

3 years ago

A missing fraction on a number line

Artyom0805 [142]

Answer:

it depends how much is the number line divided by

3 0

4 years ago

Which expression can be used to find the difference of the polynomials?(10m-6)-(7m-4)

murzikaleks [220]

Answer:

dont you already have it

7 0

3 years ago

The nth term of a quadratic sequence is:

anzhelika [568]

We can work out the first three terms of this quadratic equation with substitution.
The first term of this sequence is where n=1, we can substitute 1 in for n
y=1^2+2(1)-4
y=1+2-4
y=-1
We can do this for the next two terms as well
y=2^2+2(2)-4
y=4+4-4
y=4
For the last term:
y=3^2+2(3)-4
y=9+6-4
y=11

7 0

4 years ago