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3 years ago

Help with #7 be clear and show work !

Mathematics

2 answers:

Alex777 [14]3 years ago

4 0

Any number plus seven will always be greater than three, so just about any number would be your best bet !

Alex3 years ago

3 0

You can put any number bigger than seven because seven is always gonna be bigger than three

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Someone please help me

exis [7]

V = 1/3 l h w

V = 1/3 (12)(15)(9)
V= = 540

answer 540 m^3

8 0

3 years ago

Find the equation of the line that passes through the point (3,7) and is parallel to the line that passes through (5,5) and (5,3

Vinil7 [7]

Answer:

this is correct answer

Step-by-step explanation:

you supports to write this

7 0

3 years ago

Through: (3,3), parallel to y=8/3x+5

pshichka [43]

Answer:

(0,5)

Step-by-step explanation:

5 0

3 years ago

You are playing a card game where the total points you earn depends on theasuits of the cards you draw. The total number of poin

Semmy [17]

3a + 3b-2c-2d

a= number of hearts

b= number of diamons

c= number of clubs

d= number of spades

if you draw 3hearts, 2 diamonds, and 1 spade, just replace a=3, b=2, c=0, d=1

3(3) + 3(2)-2(0)-2(1)

Multiply each term and add the results:

9+6-0-2 = 13

13 points

4 0

1 year ago

HELPP!!!!!!!!!!!!!!!!

Anvisha [2.4K]

Answer:

Step-by-step explanation:

We'll take this step by step. The equation is

$8-3\sqrt[5]{x^3}=-7$

Looks like a hard mess to solve but it's actually quite simple, just do one thing at a time. First thing is to subtract 8 from both sides:

$-3\sqrt[5]{x^3}=-15$

The goal is to isolate the term with the x in it, so that means that the -3 has to go. Divide it away on both sides:

$\sqrt[5]{x^3}=5$

Let's rewrite that radical into exponential form:

$x^{\frac{3}{5}}=5$

If we are going to solve for x, we need to multiply both sides by the reciprocal of the power:

$(x^{\frac{3}{5}})^{\frac{5}{3}}=5^{\frac{5}{3}}$

On the left, multiplying the rational exponent by its reciprocal gets rid of the power completely. On the right, let's rewrite that back in radical form to solve it easier:

$x=\sqrt[3]{5^5}$

Let's group that radicad into groups of 3's now to make the simplifying easier:

$x=\sqrt[3]{5^3*5^2}$ because the cubed root of 5 cubed is just 5, so we can pull it out, leaving us with:

$x=5\sqrt[3]{5^2}$ which is the same as:

$x=5\sqrt[3]{25}$

8 0

3 years ago