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3 years ago

I need help with these 2 algebra questions

Mathematics

2 answers:

omeli [17]3 years ago

8 0

Answer:

question.4.

answer . b

question. 5

answer . 7m and -2m

Romashka-Z-Leto [24]3 years ago

3 0

Answer:

Step-by-step explanation:

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Find the volume of a cylinder with diameter 8 inches and height 9 1/2 inches

marta [7]

Answer:

$V=152\pi in^{2}$ in terms or pi

$V=477.52in^{2}$ simplified

Step-by-step explanation:

To solve this, we are using the Cylinder's volume formula:

$V=\pi r^{2} h$

where

$V$ is the volume of the cylinder

$r$ is the radius of the cylinder

Remember that radius is half the diameter, so we can divide the diameter of our cylinder by 2 to find its radius: $r=\frac{d}{2} =\frac{8in}{2} =4in$ . Now, 9 1/2 inches is 9 and a half inches, so $h=9.5in$ .

Replacing values:

$V=\pi (4in)^{2} (9.5in)$

$V=\pi (16in^{2})(9.5in)$

$V=152\pi in^{2}$

$V=477.52in^{2}$

We can conclude that the volume of our cylinder is $V=152\pi in^{2}$ in terms of pi, or $V=477.52in^{2}$ simplified.

4 0

3 years ago

Read 2 more answers

Use the working backward method to find the solution of the equation. 6y - 5=7 y=____

sleet_krkn [62]

Answer:

y = 2

Step-by-step explanation:

6y - 5 = 7

~Add 5 to both sides

6y = 12

~Divide 6 to both sides

y = 2

Best of Luck!

5 0

3 years ago

Read 2 more answers

What kind of triangle has 105 degress?

Black_prince [1.1K]

Answer:

obtuse angled triangle

7 0

3 years ago

point C(4,2) divides the line segment joining points A(2,-1) and B(x,y), such that AC:CB=3.1. what are the coordinates of point

White raven [17]

Answer: $B(x,y)=B(\frac{14}{3},3)$

Step-by-step explanation:

$\frac{4-2}{3} =x-4\\2=3x-12\\14=3x\\\frac{14}{3}=x\\\\\frac{2-(-1)}{3}=y-2\\3=3y-6\\9=3y\\3=y$

6 0

3 years ago

P->q is false and p is true. q is a.true b.false

kirill [66]

Answer:

Choice b. $q$ has to be false.

Step-by-step explanation:

Consider the truth table of $p \implies q$ ( $p$ implies $q$ ).

$p$ is true and $q$ is true: $p \implies q$ would be true.
$p$ is true and $q$ is false: $p \implies q$ would be false.
$p$ is false and $q$ is true: $p \implies q$ would be true.
$p$ is false and $q$ is false: $p \implies q$ would also be true.

The only combination where $p \implies q$ is false is when $p$ is true and $q$ is false. Hence, the $q\!$ here must be false.

3 0

3 years ago

I need help with these 2 algebra questions​

I need help with these 2 algebra questions