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Zolol [24]
4 years ago
12

Simplify the expression: –3(n − 5)

Mathematics
2 answers:
disa [49]4 years ago
6 0
Because of the minus sign 3 becomes -3

The answer is -3

n-5 evaluates to n-5

Multiply -3 by n-5

we multiply -3 by each term in n-5 term by term.

This is the distributive property of multiplication.

Multiply -3 and n

Multiply 1 and n

The n just gets copied along.
marusya05 [52]4 years ago
3 0

Answer:

-35+7u

Step-by-step explanation:

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True or false? help please
Andru [333]

Answer:

True

Step-by-step explanation:

Parallel lines always have the same slope, so therefore the answer is True

3 0
3 years ago
Read 2 more answers
Avery prepares 2 equal-size oranges for the bats at the zoo. one dish has 3/8 of an orange. another dish has 1/4 of an orange. W
Scorpion4ik [409]
First you convert 1/4 so it would have the same denominator as 3/8:
  1/4=2/8
 Now you compare the fractions
2/8 ? 3/8
3/8 is more, so the dish with 3/8 of an orange has more.
Hope this helps!
8 0
4 years ago
Rewrite the difference of 8 and 9 as an addition problem. 9 + 8 9 + (-8) 8 + (-9) 8 – (-9)
Naily [24]

Answer:

9+(-8)

Step-by-step explanation:

8-(-9) is subtraction

9+8 would not give you a difference

8+(-9) would give you a negative number

so 9+(-8) is the right answer

6 0
4 years ago
PLZZ HELP!! ASAP
zimovet [89]

as for my ans is C...

can you correct me if im wrongX

5 0
3 years ago
Find the derivative of the function at P 0 in the direction of A. ​f(x,y,z) = 3 e^x cos(yz)​, P0 (0, 0, 0), A = - i + 2 j + 3k
Alik [6]

The derivative of f(x,y,z) at a point p_0=(x_0,y_0,z_0) in the direction of a vector \vec a=a_x\,\vec\imath+a_y\,\vec\jmath+a_z\,\vec k is

\nabla f(x_0,y_0,z_0)\cdot\dfrac{\vec a}{\|\vec a\|}

We have

f(x,y,z)=3e^x\cos(yz)\implies\nabla f(x,y,z)=3e^x\cos(yz)\,\vec\imath-3ze^x\sin(yz)\,\vec\jmath-3ye^x\sin(yz)\,\vec k

and

\vec a=-\vec\imath+2\,\vec\jmath+3\,\vec k\implies\|\vec a\|=\sqrt{(-1)^2+2^2+3^2}=\sqrt{14}

Then the derivative at p_0 in the direction of \vec a is

3\,\vec\imath\cdot\dfrac{-\vec\imath+2\,\vec\jmath+3\,\vec k}{\sqrt{14}}=-\dfrac3{\sqrt{14}}

3 0
4 years ago
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