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3 years ago

Find the sum of 5m + 3n + p, -5p + 3n, and 2n - m.

Mathematics

2 answers:

riadik2000 [5.3K]3 years ago

8 0

Answer:

4m+8n -4p

Step-by-step explanation:

hello

you can just sum like a normal addition, just put similar terms on each other

5m + 3n +p

+3n -5p (second expression)

-m +2n ( third expression)

_____________

(5-1)m (3+3+2)n+(1-5)p = 4m+8n -4p

I hope it helps you

Have a great day

NISA [10]3 years ago

4 0

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How can I solve this question???

eduard

Answer:

144°

Step-by-step explanation:

step 1

Find the measure of the central angle of the regular pentagon

Divide 360 degrees by 5 (the number of sides)

$360\°/5=72\°$

Let

c----> the center of the pentagon

we know that

m∠ECA=72°

m∠ACB=72°

therefore

m∠ECB=m∠ECA+m∠ACB

substitute the values

m∠ECB=72°+72°=144°

The measure of angle ECB is equal to the degrees that the pentagon rotate

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3 years ago

What is greater than 0.05 but less than 0.06

ArbitrLikvidat [17]

Something between 0.05 and between 0.06 so between those to could be 0.051 0.052 0.053 0.054 0.055 0.056 0.057 0.058 0.059

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3 years ago

What are the vertical asymptote(s) of y= (x-6)/(x+8) (x-7)

White raven [17]

Answer:

x = -8 and x= 7

Step-by-step explanation:

recall that for a rational expression, the vertical asymptotes occur at x-values that causes the expression to become undefined. These occur when the denominator becomes zero.

Hence the asymptototes will occur in x-locations where the denominator , i.e

(x+8)(x-7) = 0

solving this, we get

(x+8) = 0 ----> x = -8

(x-7) = 0 ------> x = 7

hence the asymptotes occur x = -8 and x= 7

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3 years ago

Read 2 more answers

What are the zeros of f(x)=(x+5)(x-9)?

klemol [59]

Answer:

x= -5

x= 9

Step-by-step explanation:

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3 years ago

Read 2 more answers

Lori creates the following design for a T-shirt.

kakasveta [241]

<u>Solution-</u>

From the figure,

AE = 2.4

EB = 2.8

BC = 11.7

Area of rectangle 1 = 8.68 sq.in

$\Rightarrow FH \times HI=8.68$

$\Rightarrow EB \times HI=8.68$ (∵ sides of the rectangle 2)

$\Rightarrow 2.8 \times HI=8.68$

$\Rightarrow HI=3.1$

Area of Triangle 1 = 6.48 sq.in

$\Rightarrow \frac{1}{2}\times AE \times EG= 6.48$

$\Rightarrow \frac{1}{2}\times AE \times (EF+FG)= 6.48$

$\Rightarrow \frac{1}{2}\times AE \times (EF+HI)= 6.48$ (∵ sides of the rectangle 1)

$\Rightarrow EF+3.1= 5.4$

$\Rightarrow EF=2.3$

$\Rightarrow BH=2.3$ (∵ sides of the rectangle 2)

$BC = BH+HI+IC$

$\Rightarrow 11.7= 2.3+3.1+IC$

$\Rightarrow IC=6.3$

The area of Rectangle 2,

$=EB\times BH =2.8\times 2.3=6.44\ sq.in$

The area of Triangle 2,

$\frac{1}{2}\times GI \times IC=\frac{1}{2}\times EB \times IC=\frac{1}{2}\times 2.8 \times 6.3=8.82\ sq.in$

The area of the whole figure = Area of Triangle 1 + Area of rectangle 1 + Area of Triangle 2 + Area of rectangle 2

= 6.48+8.68+8.82+6.44=30.42 sq.in

6 0

3 years ago