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Georgia [21]
2 years ago
8

Please simplify this equation

Mathematics
1 answer:
yawa3891 [41]2 years ago
4 0

Answer:

2a + 8√b

Step-by-step explanation:

√9a² = 3a

√49b = 7√b

and

3a + 7√b - a + √b = 2a + 8√b

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Cari owns a horse farm and a horse trailer that can transport up to 8,000 pounds of livestock and tack. She travels with 5 horse
const2013 [10]

Answer:

5(1248 lb/horse + t lb) ≤ 8000 lb

Step-by-step explanation:

One way to do this is to find the average weight of the 5 horses:

It is:

6,240 lb

---------------- = 1248 lb/horse

  5 horses

and to this, for each horse, we must add t lbs tack.

Thus, 5(1248 lb/horse + t lb) ≤ 8000 lb

would be an appropriate inequality.

Please note that your question mentions "the following inequalities;" that means you are expected to share them!  Please do so next time.  Thanks.

5 0
3 years ago
*I WILL MARK U BRANLIEST AND GIVE 40 POINTS IF U ANSWER ALL THREE*
Gnesinka [82]

Step-by-step explanation:

1. 18(5)=90           18(6)=108

90*108= 9720

Her size of the original work was 9720 inches.

2. 1/2+1/2= 1      1/2=4 ft

The pole is 8 feet tall

3. 3 feet

3 0
3 years ago
Evaluate 3^2 + (6 − 2) ⋅ 4 − 6/3 <br><br> a. 20<br> b. 23 <br> c. 50 <br> d. 26
Sergeeva-Olga [200]
I hope this helps you

4 0
3 years ago
Read 2 more answers
PLEASE ANSWER!! GRADE 7 PRE ALGEBRA!!
NARA [144]
Divide 2106 by 27 = 78 

78 is the answer
4 0
3 years ago
Given the following right triangle, find cose, sine, tane sececsce, and cote. Do not
IgorLugansk [536]

Step-by-step explanation:

Use the Pythagorean theorem:

leg^2+leg^2=hypotenuse^2

We have

leg=4,\ hypotenuse=10

Substitute:

4^2+leg^2=10^2

16+leg^2=100             <em>subtract 16 from both sides</em>

leg^2=84\to leg=\sqrt{84}\\\\leg=\sqrt{(4)(21)}\\\\leg=\sqrt4\cdot\sqrt{21}\\\\leg=2\sqrt{21}

sine=\dfrac{opposite}{hypotenuse}\\\\cosine=\dfrac{adjacent}{hypotenuse}\\\\tangent=\dfrac{opposite}{adjacent}\\\\cotangent=\dfrac{adjacent}{opposite}\\\\secant=\dfrac{hypotenuse}{adjacent}\\\\cosecant=\dfrac{hypotenuse}{opposite}

Substitute:

hypotenuse=10,\ opposite=4,\ adjacent=2\sqrt{21}

\sin\theta=\dfrac{4}{10}=\dfrac{2}{5}\\\\\cos\theta=\dfrac{2\sqrt{21}}{10}=\dfrac{\sqrt{21}}{5}\\\\\tan\theta=\dfrac{4}{2\sqrt{21}}=\dfrac{2}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{2\sqrt{21}}{21}\\\\\cot\theta=\dfrac{2\sqrt{21}}{4}=\dfrac{\sqrt{21}}{2}\\\\\sec\theta=\dfrac{10}{2\sqrt{21}}=\dfrac{5}{\sqrt{21}}\cdot\dfrac{\sqrt{21}}{\sqrt{21}}=\dfrac{5\sqrt{21}}{21}\\\\\csc\theta=\dfrac{10}{4}=\dfrac{5}{2}

5 0
3 years ago
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