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Anit [1.1K]
3 years ago
10

The sum of 10 and a product of 7 and a number

Mathematics
1 answer:
NeX [460]3 years ago
8 0
The answer to your question, would be 7x + 10.

Hopefully this helps!
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Right Triangles I been having trouble with gemotrey lately ​
photoshop1234 [79]

Answer:

x = 11√3

y = 22

Step-by-step explanation:

In a 30-60-90 triangle, the hypotenuse is twice the shortest side.

y = 22

The long side is √3 times the short side.

x = 11√3

If you wish to use trigonometric functions:

sin 30° = 11 / y

1/2 = 11 / y

y = 22

tan 30° = 11 / x

1 / √3 = 11 / x

x = 11√3

4 0
3 years ago
Help please? 10 points. Correctly
Romashka [77]

From the Figure :

Point A is (-3 , -3)

Point B is (6 , 6)

We know that, The Mid Point of Two Points (x₁ , y₁) and (x₂ , y₂) is given by :

\mathsf{\implies [(\frac{x_1 + x_2}{2})\;,\;(\frac{y_1 + y_2}{2})]}

\mathsf{\implies Midpoint\;of\;Line\;AB\;is\;[(\frac{-3 + 6}{2})\;,\;(\frac{-3 + 6}{2})]}

\mathsf{\implies Midpoint\;of\;Line\;AB\;is\;[(\frac{3}{2})\;,\;(\frac{3}{2})]}

\mathsf{\implies Midpoint\;of\;Line\;AB\;is\;(1.5,1.5)}

4 0
3 years ago
It’s due tomorrow plz HELP
Vikentia [17]

1)

\dfrac{3}{4}  + 5 -  \dfrac{1}{2}  = \dfrac{3}{4}  +  \dfrac{20}{4} -  \dfrac{2}{4}  =  \dfrac{21}{4}  = 5 \dfrac{1}{4}

2)

\dfrac{2}{3}  +  {2}^{3}  -  \dfrac{1}{3}  =  \dfrac{2}{3}  +  \dfrac{24}{3}  -  \dfrac{1}{3}  =  \dfrac{25}{3}  = 8 \dfrac{1}{3}

3)

- 15 + 6x - 19 - 20x =  - 14x - 34

4)

- 7x - 2 - 9y + 10x = 3x - 9y - 2

5)

14 + x = 18 \\ x = 18 - 14 \\ x = 4

6)

- 5x =  - 15 \\ x =  - 15 \div ( - 5) \\ x = 3

7)

- 2x + 8 = 18 \\  - 2x = 18 - 8 \\  - 2x = 10 \\ x = 10 \div ( - 2) \\ x =  - 5

8)

3x + 18 - 8x =  - 17 \\ 3x - 8x =  - 17 - 18 \\  - 5x =  - 35 \\ x =  - 35 \div ( - 5) \\ x = 7

7 0
3 years ago
How do you solve this? 18 " n=21
dalvyx [7]
18 ^{n} =21

log(18) ^{n}=log(21)

n*log(18)=log(21)

n= \frac{log(21)}{log(18)}

n=1.053
6 0
3 years ago
Which type of statement must be proven in geometry? A. axiom
Leya [2.2K]
D a theorem needs to be proven
8 0
3 years ago
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