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3 years ago

Solve 9x = 27 1/3 2/3 3/2

Mathematics

1 answer:

Rina8888 [55]3 years ago

6 0

Answer:

the answer is x=3

Step-by-step explanation:

divide by 9 on each side of = sign

9/9 and 27/9=3

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10. Melissa has $20 to buy bagels and juice for her class. box of bagels bottle of juice 56.13, including tax $2.08, including t

adelina 88 [10]

The inequality for the maximum number of bottles of juice she can buy

6.13+ 2.08b <20.00

Option B

Given

Melissa has $20 to buy bagels and juice for her class

Box of bagels : 6.13 including tax

Bottle of juice: 2.08 including tax

She will buy only one box of bagels that is for 6.13

Let 'b' be the number of bottles of juice she can buy

Total cost should be less than 20 dollars

6.13(one bagels) +2.08(bottles of juices)<20

So the inequality becomes

$6.13+2.08b$

Learn more : brainly.com/question/1836165

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2 years ago

What would 135 represent in the data set below?

Citrus2011 [14]

C. The outlier because it doesn’t fit with the rest

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3 years ago

A fire is sighted Due West of Lookout A. The bearing of the fire from Lookout B, 6.5 miles due south of A, is North 35°34'W. How

KonstantinChe [14]

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4 0

3 years ago

PLEASE ANSWER THISSSSSSSSSSSSS

Setler [38]

Answer:

D. 19.5 Ft.

Step-by-step explanation:

Formula: B · H = A

Substitution: 3.25 · 2 = 6.5

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8 0

3 years ago

Read 2 more answers

Can you guys help me out on this? I'm still learning sign, cosign, and tangent :)

Yakvenalex [24]

Answer:

$\sin d = \frac{4}{7}$ ; $\sin e = \frac{\sqrt{33} }{7}$

$\cos d = \frac{\sqrt{33} }{7}$ ; $\cos e = \frac{4}{7}$

$\tan d = \frac{4}{\sqrt{33} }$ ; $\tan e = \frac{\sqrt{33} }{4}$

Step-by-step explanation:

For a right angled triangle with one of its angle α (alpha) :-

$\sin \alpha = \frac{Side \: opposite \: to \: \alpha }{Hypotenuse \: of \: the \: triangle}$

$\cos \alpha = \frac{Side \: adjacent \: to \: \alpha }{Hypotenuse \: of \: the \: triangle}$

$\tan \alpha = \frac{Side \: opposite \: to \: \alpha }{Side \: adjacent \: to \: \alpha }$

__________________________________________________

According to the question ,

1) When α (alpha) = d

$\sin d = \frac{4}{7}$

$\cos d = \frac{\sqrt{33} }{7}$

$\tan d = \frac{4}{\sqrt{33} }$

2) When α (alpha) = e

$\sin e = \frac{\sqrt{33} }{7}$

$\cos e = \frac{4}{7}$

$\tan e = \frac{\sqrt{33} }{4}$

3 0

3 years ago