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PSYCHO15rus [73]
3 years ago
12

PLEASE HELP IM DUMB AND I NED THIS ASAP

Mathematics
1 answer:
netineya [11]3 years ago
6 0

Width (7.8) x length (7.8) x width (7.8) = 474.552

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Which fractions 1/3, 5/6, 3/4, 3/8 are closer to 0 than to 1?
aev [14]

Answer:

1/3, 3/8 are closer to 0

Step-by-step explanation:

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Marisol buys 3 pounds of cheese and 3 pounds of sausage for a total cost of $36. The sausage costs $2.00 less per pound than the
Andreas93 [3]

Answer:

The answer is 1pound of cheese is 7$ and the sausage is 5$

Step-by-step explanation:

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3 years ago
Without calculating, which has a bigger volume. A cube that has a length, width, and height of 18 m. Or a sphere with a radius o
Evgesh-ka [11]

Weird. A period appears above this... huh.

Answer:

[Th]e cube has a greater value.

Step-by-step explanation:

What the word problem really wants us to get [is ]the question of 'Which is greater, A=6a^2 when 'a' [is] 18 or A=4\pir^2 when r = 9? And here's how to solve that.

Starting with the[ c]ube we have A=6a^2. A bit t[o]o simple, right?

A=6(18)^2 Substitute numbers.

A=6(324) Solve ex[p]onents.

A=1944 Mult[i]ply.

So w[e] know that the cube is 1944 meters cube[d ] in area. But what about the more [f]ormidable sphere? Fo[r] it we need a slightly m[o]re co[m]plicated formula, A=4\pir^2. However, instead of using the real pi I will be rounding to 3.14, since we have no calculator so anything more would take way too long and fry your[ bra]in.

A=4(3.14)(9)^2 Subst[i]tute numbers.

A=4(3.14)(81) Solve expone[n]ts.

A=12.56(81) Multip[ly].

A=1017.36 Multiply again[.]

Now, since I'm sure all of us can count, we know that 1944 is greater than 1017.36. Or in other words, the cube is bigger than the sphere.

And PLEASE don't copy this guys. Make your own iteration. Change it up!

3 0
3 years ago
Find the solution set of the inequality
Brilliant_brown [7]

<u>Answer:</u>

\boxed{\textsf{ Hence the solution set is$ \bf{\bigg( \dfrac{33}{4},\infty \bigg) }$}}.

<u>Step-by-step explanation:</u>

A inequality is given to us and we need to find the solution set. So the given inequality to us is ,

  • -4x + 35 >2

\bf\implies -4x + 35 > 2 \\\\\bf\implies -4x > 2-35 \\\\\bf \implies -4x > -33 \\\\\bf \implies x >\dfrac{-33}{-4}\\\\\bf \implies x > \dfrac{33}{4} \\\\\bf\implies \boxed{\pink{\bf x \in \bigg(\dfrac{33}{4}, \infty \bigg) }}

<h3><u>★</u><u> </u><u>Hence </u><u>the </u><u>solution</u><u> </u><u>set </u><u>is </u><u>x </u><u>€</u><u> </u><u>(</u><u> </u><u>3</u><u>3</u><u>/</u><u>4</u><u> </u><u>,</u><u> </u><u>∞</u><u> </u><u>)</u><u>.</u></h3>
3 0
3 years ago
Use the rules of exponents to simplify the expressions. Match the expression with its equivalent value.
Lelechka [254]

Answer:

1) \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4} = \frac{1}{32}

3) (-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}

4) \frac{2}{2^{-4}} = 32

Step-by-step explanation:

1) \frac{(-2)^{-5}}{(-2)^{-10}}

Solving using exponent rule: a^{-m}=\frac{1}{a^m}

\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32

So, \frac{(-2)^{-5}}{(-2)^{-10}}=-32

2) 2^{-1}.2^{-4}

Using the exponent rule: a^m.a^n=a^{m+n}

We have:

2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}

We also know that: a^{-m}=\frac{1}{a^m}

Using this rule:

2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}

So, 2^{-1}.2^{-4} = \frac{1}{32}

3) (-\frac{1}{2} )^3.(-\frac{1}{2} )^2

Solving:

(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}

So, (-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}

4) \frac{2}{2^{-4}}

We know that: a^{-m}=\frac{1}{a^m}

\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32

So, \frac{2}{2^{-4}} = 32

3 0
3 years ago
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