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2 years ago

Find the difference and simplify 3/4 - 2/8

Mathematics

1 answer:

Maksim231197 [3]2 years ago

3 0

Answer:

1/2

Step-by-step explanation:

(6/8)-(2/8)=(4/8)

(4/8) = 1/2

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List the sides of each triangle from shortest to greatest

Lelechka [254]

In every triangle, the largest side, is opposite the largest angle.

Also, the smallest side, is opposite the smallest angle.

a. $70^o\ \textgreater \ 60^o\ \textgreater \ 50^o$ so $|RQ|\ \textgreater \ |PR|\ \textgreater \ |PQ|$ ,

because RQ is opposite angle P, the largest,
PR is opposite angle Q , the middle
and PQ is opposite angle R, the smallest.

Greatest side: RQ

b. The sum of the measures of the interior angles of any triangle is 180°,

so

$m(A)+m(B)+m(C)=180^o\\\\50^o+45^o+x=180^o\\\\x=180^o-50^o-45^o=85^o$

thus, m(C) is the largest angle. This means the largest side is the side AB.

Answers:

a) RQ
b)AB

8 0

3 years ago

Find the domain and range of the given relation (6,3), (-7,-3), (-4,1), (1,-4)

lions [1.4K]

Answer:

Step-by-step explanation:

The domain are all the x values

(-7,-3), (-4,1), (1,-4), (6,3)

x here is from -7 to 6

domain ∈ [-7, 6]

The range are all the y values

(1,-4), (-7,-3), (-4,1), (6,3)

y here is from -4 to 3

range ∈ [-4, 3]

3 0

2 years ago

I need help on this problem, I dont know how to do it, I dont know if its about finding area or anything but its for Geometry. I

miskamm [114]

Answer:

Step-by-step explanation:

<B=90-47=43

AB/BC=sin 47

BC=AB/sin 47=18/0.731=24.6 nearly

AC/AB=Tan 43

AC=18 tan 43=16.8 nearly

4 0

3 years ago

77. the volume of a cube is increasing at a rate of <img src="https://tex.z-dn.net/?f=10%20%5Cmathrm%7B~cm%7D%5E%7B3%7D%20%2F%20

Colt1911 [192]

Answer:

$\displaystyle \frac{4}{3}\text{cm}^2/\text{min}$

Step-by-step explanation:

Given

$\displaystyle \frac{dV}{dt}=10\:\text{cm}^3/\text{min}\\ \\V=s^3\\\\SA=6s^2\\\\\frac{d(SA)}{dt}=?}\:;s=30\text{cm}$

Solution

(1) Find the rate of the cube's edge length with respect to time at s=30:

$\displaystyle V=s^3\\\\\frac{dV}{dt}=3s^2\frac{ds}{dt}\\ \\10=3(30)^2\frac{ds}{dt}\\ \\10=3(900)\frac{ds}{dt}\\\\10=2700\frac{ds}{dt}\\\\\frac{10}{2700}=\frac{ds}{dt}\\\\\frac{ds}{dt}=\frac{1}{270}\text{cm}/\text{min}$

(2) Find the rate of the cube's surface area with respect to time at s=30:

$\displaystyle SA=6s^2\\\\\frac{d(SA)}{dt}=12s\frac{ds}{dt}\\ \\\frac{d(SA)}{dt}=12(30)\biggr(\frac{1}{270}\biggr)\\\\\frac{d(SA)}{dt}=\frac{360}{270}\biggr\\\\\frac{d(SA)}{dt}=\frac{4}{3}\text{cm}^2/\text{min}$

Therefore, the surface area increases when the length of an edge is 30 cm at a rate of $\displaystyle \frac{4}{3}\text{cm}^2/\text{min}$ .

6 0

2 years ago

Tanisha solved the equation below by graphing a system of equations. log3 5x = log5 ( 2x + 8). What system of equations could be

Marta_Voda [28]

Answer:

The set

$y=log_3 (5x)$

$y=log_5 (2x+3)$

Step-by-step explanation:

We separate the left and right side of the equation, to create two new equations:

(1) $y=log_3 (5x)$

(2). $y=log_5 (2x+3)$

and then to solve Tanisha's original equation, we find the solution to the system that we got above.

The first step in solving the above system of equations is to take the value for $y$ from equation(1), and substitute it into equation(2) which gives us:

$y=log_5 (2x+3)\\\\ \boxed{log_3(5x) = log_5 (2x+3)}$

which is the same equation that Tanisha set out to solve.

5 0

3 years ago