All categories

2 years ago

3(b − 6) = 9 b = The test is timed

Mathematics

2 answers:

marusya05 [52]2 years ago

6 0

Answer:

b = 9

Step-by-step explanation:

3(b − 6) = 9

3b - 18 = 9

3b = 9 + 18

3b = 27

b = 9

RoseWind [281]2 years ago

6 0

Answer:

b = 9

Step-by-step explanation:

$3(b - 6) = 9$

$3(b - 6) \div 3 = 9 \div 3$

$b - 6 = 9 \div 3$

$b - 6 = 3$

$b - 6 + 6 = 3 + 6$

$b = 3 + 6$

$b = 9$

You might be interested in

Find the size of the final unknown interior angle in a polygon whose other interior angles are:

qaws [65]

Answer:

Step-by-step explanation:(2ñ-4)90

(7×2-4)90

=900°

900-(162+125+148+105+98+115)

900-753

=147°

3 0

2 years ago

Randy is painting part of a wall, excluding the two corners shown in white. a gallon of white paint will cover 100 sqaure feet.

uysha [10]

Randy needs to purchase 1.49 Gallons of paint to cover 100 sq. feet wall.

What is the area of rectangle?

The area of a rectangle is the product of its length and width. That is, A = l x w where l is the length and w is the width of the rectangle.

Here, Required Paint = $20 X \frac{37}{4} - 2 X \frac{1}{2}X6 X 6\\$

= 5 X 37 - 36

= 185 - 36

= 149

= 1.49 gallons

Thus, Randy needs to purchase 1.49 Gallons of paint to cover 100 sq. feet wall.

Learn more about Area of Rectangle from :

brainly.com/question/20693059

#SPJ1

4 0

2 years ago

F(t)=t^2+19t+60<br> What are the zeros of the function and what is the vertex of the parabola?

SVEN [57.7K]

Answer: $-4,-15;\ \left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$

Step-by-step explanation:

Given

Function is $F(t)=t^2+19+60$

Zeroes of the function are

$t^2+19t+60=0\\\\\Rightarrow t=\dfrac{-19\pm\sqrt{19^2-4\times 1\times 60}}{2\times 1}\\\\\Rightarrow t=\dfrac{-19\pm \sqrt{121}}{2}\\\\\Rightarrow t=\dfrac{-19\pm 11}{2}\\\\\Rightarrow t=-4,-15$

Using completing the square method

$y=t^2+2\times \dfrac{19}{2}t+\dfrac{19^2}{2^2}-\dfrac{19^2}{2^2}+60\\\\y=\left(t+\dfrac{19}{2}\right)^2+60-\dfrac{361}{4}\\\\y=\left(t+\dfrac{19}{2}\right)^2-\dfrac{121}{4}\\\\y+\dfrac{121}{4}=\left(t+\dfrac{19}{2}\right)^2\\\\\left(y-(-\dfrac{121}{4}\right)=\left(t-(-\dfrac{19}{2})\right)^2\\\\\text{The vertex is }\left(-\dfrac{19}{2},-\dfrac{121}{4}\right)$