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tekilochka [14]

3 years ago

12

Find the area of a right triangle with side lengths 12 cm, 35 cm, and 37 cm. Then find the length of the altitude drawn to the h

ypotenuse

1 answer:

HACTEHA [7]3 years ago

7 0

For area multiply the numbers

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Subtract -7n +2 from 2n - 1

Irina18 [472]

2n - 1 - -7n + 2 You combine like terms with subtracting
2n - 7n = -5n
-1 - 2 = -3
So -7n + 2 - 2n - 1 = -5n - 3

4 0

3 years ago

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Lelechka [254]

Answer:

5 quartz would make the most sense, hope this helps

Step-by-step explanation:

7 0

2 years ago

Add the following fractions. See the image below

LenKa [72]

Answer:

$\dfrac{19}{840}$

Step-by-step explanation:

The given fraction is:

$\dfrac{1}{168}+\dfrac{3}{180}$

We need to solve it.

The LCM of 168 and 180 is 2520.

So,

$\dfrac{1}{168}+\dfrac{3}{180}=\dfrac{15+3\cdot14}{2520}\\\\=\dfrac{19}{840}$

So, the required answer is equal to $\dfrac{19}{840}$ .

3 0

2 years ago

What is the value of w? 28=4(w−3)

sveticcg [70]

Answer:

w=10

Step-by-step explanation:

1. Divide 4 on both sides to cancel out the 4 on the right side since it is an equation so it has to remain “equal”/ balanced.

2. You get 7=w-3.

3. Add 3 on both sides.

4.You get w=10.

7 0

3 years ago

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Solve (x + 4)2 – 3(x + 4) – 3 = 0 using substitution. u = Select the solution(s) of the original equation.

bezimeni [28]

Answer:

$\dfrac{-5-\sqrt{21}}{2},\ \dfrac{-5+\sqrt{21}}{2}$

Step-by-step explanation:

For the equation $(x+4)^2-3(x+4)-3=0$ use the substitution $u=x+4.$ Then the equation will take look

$u^2-3u-3=0.$

Solve this quadratic equation:

$D=(-3)^2-4\cdot 1\cdot (-3)=9+12=21,\\ \\u_{1,2}=\dfrac{-(-3)\pm \sqrt{21}}{2}=\dfrac{3\pm\sqrt{21}}{2}.$

Thus,

$x+4=\dfrac{3-\sqrt{21}}{2}\text{ or }x+4=\dfrac{3+\sqrt{21}}{2},\\ \\x_1=\dfrac{3-\sqrt{21}}{2}-4=\dfrac{-5-\sqrt{21}}{2}\text{ or }x_2=\dfrac{3+\sqrt{21}}{2}-4=\dfrac{-5+\sqrt{21}}{2}.$

8 0

3 years ago

Read 2 more answers