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3 years ago

Solve the equation 3|x – 2|+2 = 17

Mathematics

1 answer:

grigory [225]3 years ago

5 0

Answer:

-3, 7 = x

Step-by-step explanation:

3|x – 2| + 2 = 17

- 2 - 2

________________

3|x - 2| = 15

_______ ___

3 3

|x - 2| = 5

$x - 2 = -5 \: \: \: \: \: \: \: \: \: \: \: \: \: x - 2 = 5$

+ 2 + 2 + 2 + 2

_______________ ____________

x = -3 x = 7

I am joyous to assist you anytime.

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Kaylis [27]

4*7=28
5y=28
divide 5 to 28
y=5.6

7 0

3 years ago

Name:

qaws [65]

Answer:

Width of the rectangle = 5 inches

Length of the rectangle = 20 inches

Step-by-step explanation:

Let us assume the width of the rectangle = m in

So, the length of the rectangle = 4 (Width) = (4 m) in

Perimeter of the rectangle = 50 inches

Now, Perimeter of the Rectangle = 2 (LENGTH + WIDTH)

⇒ 2( m + 4m) = 50

or, 2(5m) = 50

⇒ 10m = 50 or, m = 50/10 = 5

⇒ m = 5

So, the width of the rectangle = 5 inches

and the length of the rectangle = 4 m = 4 x 5 = 20 inches

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2 years ago

What conclusions can be drawn about finding the quotient in scientific notation? Check all that apply.

irga5000 [103]

Answer:

B.The exponent of the solution is –12, the difference of the original exponents.

C.The coefficient of the solution must be greater than or equal to one but less than 10.

D.The quotient is 3.0 ×

Step-by-step explanation:

7 0

1 year ago

Read 2 more answers

Solve the following and explain your steps. Leave your answer in base-exponent form. (3^-2*4^-5*5^0)^-3*(4^-4/3^3)*3^3 please st

Naily [24]

Answer:

$\boxed{2^{\frac{802}{27}} \cdot 3^9}$

Step-by-step explanation:

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Texting in Latex is much more clear and depending on the question, just writing down without it may be confusing or ambiguous. Be together with Latex! （＊＾Ｕ＾）人（≧Ｖ≦＊）/

$(3^{-2} \cdot 4^{-5} \cdot 5^0)^{-3} \cdot (4^{-\frac{4}{3^3} })\cdot 3^3$

Note that

$\boxed{a^{-b} = \dfrac{1}{a^b}, a\neq 0 }$

The denominator can't be 0 because it would be undefined.

So, we can solve the expression inside both parentheses.

$\left(\dfrac{1}{3^2} \cdot \dfrac{1}{4^5} \cdot 5^0 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{3^3} } }\right)\cdot 3^3$

Also,

$\boxed{a^{0} = 1, a\neq 0 }$

$\left(\dfrac{1}{9} \cdot \dfrac{1}{1024} \cdot 1 \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

Note

$\boxed{\dfrac{1}{a} \cdot \dfrac{1}{b}= \frac{1}{ab} , a, b \neq 0}$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{1}{4^{\frac{4}{27} } }\right)\cdot 27$

$\left(\dfrac{1}{9216} \right)^{-3} \cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left( \dfrac{1}{\left(\dfrac{1}{9216}\right)^3} \right)\cdot \left(\dfrac{27}{4^{\frac{4}{27} } }\right)$

Note

$\boxed{\dfrac{1}{\dfrac{1}{a} } = a}$

$9216^3\cdot \left(\dfrac{27}{4^{\frac{4}{9} } }\right)$

$\left(\dfrac{ 9216^3\cdot 27}{4^{\frac{4}{27} } }\right)$

Once

$9216=2^{10}\cdot 3^2 \implies 9216^3=2^{30}\cdot 3^6$

$\boxed{(a \cdot b)^n=a^n \cdot b^n}$

And

$4^{\frac{4}{27}} = 2^{\frac{8}{27}$

We have

$\left(\dfrac{ 2^{30} \cdot 3^6\cdot 27}{2^{\frac{8}{27} } }\right)$

Also, once

$\boxed{\dfrac{c^a}{c^b}=c^{a-b}}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27$

$30-\dfrac{8}{27} = \dfrac{30 \cdot 27}{27}-\dfrac{8}{27} =\dfrac{802}{27}$

$2^{30-\frac{8}{27}} \cdot 3^6\cdot 27 = 2^{\frac{802}{27}} \cdot 3^6 \cdot 3^3$

$2^{\frac{802}{27}} \cdot 3^9$

4 0

2 years ago

Select the correct answer.

lorasvet [3.4K]

Add 5 to both sides of the equation.
x=-14+5
Add -14 and 5.
x=-9

8 0

2 years ago