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3 years ago

Pls Help Soon as possible

Mathematics

1 answer:

guajiro [1.7K]3 years ago

7 0

The first you can multiply first equation by 2, you will get 16x - 12y = - 40, then add first equation to second

16x - 12y - 16x + 7y = - 40 + 30
-5y = - 10
Y= 2;
Then put y to first equation, because it easier.
8x - 2*6=-20
8x = - 8
X= - 1

Answer is x = - 1, y = 2.

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A college student earned $7000 during summer vacation working as a waiter in a popular restaurant. The student invested part of

Lemur [1.5K]

.09X+.08(7,000-X)=588
.09X+560-.08X=588
.01X=588-560
.01X=28
X=28/.01

X=2,800 AMOUNT INVESTED @ 9%.
7,000-2,800=4,200 AMOUNT INVESTED @ 8%
PROOF:
.09*2,800+.08*4,200=588
252+336=588
588=588

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4 years ago

0.0084 scientific notation

kogti [31]

Answer:

8.4*10^-3

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4 years ago

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Sec (90 -A) sinA = cot (90-A). tan( 90- A)

djyliett [7]

Step-by-step explanation:

sec(90-A) . Sin A = cot (90-A) . tan(90-A)

cosec X sinA = tanA X cotA

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1=1

Hence proved

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3 years ago

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today each cookie costs 0.75 less than the normal price. pight now if you buy 7 of them it will only cost you 2.80 . writw an eq

lbvjy [14]

The normal price of each cookie is $1.15 derived from the equation, c =today price of each cookie + 0.75

Solution:

Given, Today each cookie costs 0.75 less than the normal price.

Right now if you buy 7 of them it will only cost you 2.80

We have to write an equation to determine the normal price of each cookie(c)

And, we are given that, today 7 cookies cost 2.80

Let the cost of one cookie today be “n”

Then we get,

Today 7 cookies ⇒ 2.8

Then, 1 cookie ⇒ n

By criss cross multiplication we get,

$\begin{array}{l}{7 \times n=2.8 \times 1} \\\\ {\rightarrow 7 n=2.8} \\\\ {\rightarrow n=0.4}\end{array}$

Which means that, each cookie costs $0.4 today

Now, we know that,

normal price of each cookie, c = today price of each cookie + $ 0.75

c = 0.4 + 0.75

c= 1.15

Hence, the normal cost of each cookie is $1.15 derived from the equation, c = today price of each cookie + 0.75

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3 years ago

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g100num [7]

Answer:

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3 years ago