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stealth61 [152]
3 years ago
8

What is the GCF of 77 and 56?

Mathematics
2 answers:
AveGali [126]3 years ago
5 0
The GCF is seven. 7*11 and 7*8
faltersainse [42]3 years ago
5 0
The gcf of 77 and 56 is 7
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The table shows the altitudes of four different cities. What is the correct way to arrange the altitudes in decreasing order?
ki77a [65]

Answer:

A. 58, 13,-35,-51

Step-by-step explanation:

Start with the greatest positive number, 58. Then use the next greatest positive number, 13. When it comes to negative numbers, the number with the smaller absolute value is greater.

|-35| = 35

|-53| = 53

-35 has a smaller absolute value than -53, so -35 is greater than -53.

Answer: A. 58, 13,-35,-51

7 0
3 years ago
Read 2 more answers
What is the anwser -13-4x=x+7
Luden [163]

-13-4x=x+7

+13         +13

-4x=x+20

-x    -x

-5x=20

/-5   /-5

x=-4

---

hope it helps

7 0
2 years ago
Please someone help me to prove this..​
Pachacha [2.7K]

Answer:  see proof below

<u>Step-by-step explanation:</u>

Use the following Sum to Product Identities:

\sin x+\sin y=2\sin \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\sin x-\sin y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=-2\sin \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)

<u>Proof LHS → RHS</u>

\text{LHS:}\qquad \qquad \qquad \dfrac{\sin 5-\sin 15+\sin 25 - \sin 35}{\cos 5-\cos 15- \cos 25 + \cos 35}

\text{Reqroup:}\qquad \qquad \qquad \dfrac{(\sin 25+\sin 5)-(\sin 35 + \sin 15)}{(\cos 35+\cos 5)-(\cos 25 + \cos 15)}

\text{Sum to Product:}\quad \dfrac{2\sin \bigg(\dfrac{25+5}{2}\bigg)\cos \bigg(\dfrac{25-5}{2}\bigg)-2\sin \bigg(\dfrac{35+15}{2}\bigg)\cos \bigg(\dfrac{35-15}{2}\bigg)}{2\cos \bigg(\dfrac{25+15}{2}\bigg)\cos \bigg(\dfrac{25-15}{2}\bigg)-2\cos \bigg(\dfrac{35+5}{2}\bigg)\cos \bigg(\dfrac{35-5}{2}\bigg)}\text{Simplify:}\qquad \qquad \dfrac{2\sin 15\cos 10-2\sin 25\cos 10}{2\cos 20\cos 15-2\cos 20\cos 5}

\text{Factor:}\qquad \qquad \dfrac{2\cos 10(\sin 15-\sin 25)}{2\cos 20(\cos 15-\cos 5)}

\text{Sum to Product:}\qquad \dfrac{\cos 10\bigg[2\cos \bigg(\dfrac{15+25}{2}\bigg)\sin \bigg(\dfrac{15-25}{2}\bigg)\bigg]}{\cos 20\bigg[-2\sin \bigg(\dfrac{15+5}{2}\bigg)\sin \bigg(\dfrac{15-5}{2}\bigg)\bigg]}

\text{Simplify:}\qquad \qquad \dfrac{\cos 10[2\cos 20\sin (-5)]}{\cos 20[-2\sin 10\sin 5]}\\\\\\.\qquad \qquad \qquad =\dfrac{-2\cos10 \cos 20 \sin 5}{-2\sin 10 \cos 20 \sin 5}\\\\\\.\qquad \qquad \qquad =\dfrac{\cos 10}{\sin 10}\\\\\\.\qquad \qquad \qquad =\cot 10

LHS = RHS:  cot 10 = cot 10   \checkmark

8 0
2 years ago
Solve the problem using ELIMINATION<br> 20x + 2y = 18 10x + y = 9
sesenic [268]

Answer:

Infinitely many solutions.

Step-by-step explanation:

Multiply the second equation by -2, then add the equations together

(20x+2y=18)

−2(10x+y=9)  

20x+2y=18

−20x−2y=−18

Add these equations to eliminate x

0=0

Infinitely many solutions.

4 0
2 years ago
Will give brainliest
jeyben [28]

Answer:

<h2> <em><u>384</u></em></h2>

Step-by-step explanation:

<em><u>Given</u></em><em><u>, </u></em>

Dimensions of the rectangular prism = 9ft, 12ft and 4ft

<em><u>Therefore</u></em><em><u>, </u></em>

Surface area of the rectangular prism

= 2( lb + bh + lh)

= 2(12 \times 9 + 9 \times 4 + 12 \times 4) {ft}^{2}

= 2(108 + 36 + 48) {ft}^{2}

= 2 \times 192 {ft}^{2}

= 384 {ft}^{2}

<em><u>Hence</u></em><em><u>,</u></em>

<em><u>Surface</u></em><em><u> </u></em><em><u>area</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>rectangular</u></em><em><u> </u></em><em><u>prism</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>384</u></em><em><u> </u></em><em><u>sq</u></em><em><u>.</u></em><em><u> </u></em><em><u>ft</u></em><em><u> </u></em><em><u>(</u></em><em><u>Ans</u></em><em><u>)</u></em>

3 0
3 years ago
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