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Citrus2011 [14]
2 years ago
12

What is the answer to 4/5w = 8

Mathematics
2 answers:
Ahat [919]2 years ago
8 0

Answer:7 1/5

Step-by-step explanation:

-4/5 on each side

'and you get w= 7 1/5

Leni [432]2 years ago
4 0

Answer:

w = 10

Step-by-step explanation:

\frac{4}{5}w=8~(Given)\\\\\frac{5}{4}(\frac{4}{5}w)=\frac{5}{4}(8)~(Multiply~5/4~on~both~sides)\\\\w=10~(Simplify)

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For the following inequality, indicate whether the boundary line should be dashed or solid. x ≤ 5
lukranit [14]
Answer: The boundary lien should be solid because the inequality indicates that the value “5” is included. Remember: “greater than or equal to; less than or equal to,” as solid lines. While “<;>” are dashed lines.
6 0
3 years ago
Whoever answers first gets branliest Emily drops a stone from rest off a high bridge into a river. What is the speed of the ston
Nataliya [291]

Answer: B: -49

Step-by-step explanation:

8 0
4 years ago
Read 2 more answers
Prove the divisibility:<br><br>45^45·15^15 by 75^30
garri49 [273]

Answer:

3^{75}.

Step-by-step explanation:

We have been an division problem: \frac{45^{45}*15^{15}}{75^{30}}.

We will simplify our division problem using rules of exponents.

Using product rule of exponents (a*b)^n=a^n*b^n we can write:

45^{45}=(9*5)^{45}=9^{45}*5^{45}

15^{15}=(3*5)^{15}=3^{15}*5^{15}

75^{30}=(15*5)^{30}=15^{30}*5^{30}

Substituting these values in our division problem we will get,

\frac{9^{45}*5^{45}*3^{15}*5^{15}}{15^{30}*5^{30}}

Using power rule of exponents a^n*a^m=a^{n+m} we will get,

\frac{9^{45}*5^{(45+15)}*3^{15}}{15^{30}*5^{30}}

\frac{9^{45}*5^{60}*3^{15}}{15^{30}*5^{30}}

Using product rule of exponents (a*b)^n=a^n*b^n we will get,

\frac{(3*3)^{45}*5^{60}*3^{15}}{(3*5)^{30}*5^{30}}

\frac{3^{45}*3^{45}*5^{60}*3^{15}}{3^{30}*5^{30}*5^{30}}

Using power rule of exponents a^n*a^m=a^{n+m} we will get,

\frac{3^{(45+45+15)}*5^{60}}{3^{30}*5^{(30+30)}}

\frac{3^{105}*5^{60}}{3^{30}*5^{60}}

\frac{3^{105}}{3^{30}}

Using quotient rule of exponent \frac{a^m}{a^n}=a^{m-n} we will get,

\frac{3^{105}}{3^{30}}=3^{105-30}

3^{105-30}=3^{75}

Therefore, our resulting quotient will be 3^{75}.

7 0
3 years ago
Please help me <br> 10 points
Nonamiya [84]

Answer:

  see below

Step-by-step explanation:

The following definitions apply:

  if p, then q . . . . conditional statement

  if q, then p . . . . converse

  if ~p then ~q . . . inverse

  if ~q then ~p . . . contrapositive

You have ...

  p = "it is October 31"

  q = "it is Halloween"

(a) The <em>converse</em> is ...

  If it is Halloween, then it is October 31.

(b) The <em>inverse</em> is ...

  If it is not October 31, then it is not Halloween.

(c) The <em>contrapositive</em> is ...

  if it is not Halloween, then it is not October 31.

4 0
3 years ago
What dimensions of the box maximize the volume of the​ box?<br><br> please help thank you so much!
bonufazy [111]

so, let's recall that the volume of a rectangular prism, namely a box, is Lwh, just the product of its dimensions, length, width and height.

now, check the 1st picture below. we know the original paperboard is a 16x14, so the width, as you see in the picture is w = 8 - x, its length is L = x, and its height is 14 -(x/2) - (x/2).


\bf V(x)=\stackrel{L}{(x)}\stackrel{w}{(8-x)}\stackrel{h}{\left( 14-\cfrac{x}{2}-\cfrac{x}{2} \right)}\implies V(x)=(8x-x^2)(14-x) \\\\\\ V(x)=112x-8x^2-14x^2+x^3\implies V(x)=x^3-22x^2+112x


now, let's find the critical points, by setting the derivative to 0, and then we'll do a first-derivative test.


\bf \cfrac{dV}{dx}=3x^2-44x+112\implies 0=3x^2-44x+112 \\\\\\ \textit{plugging those values in the quadratic formula}\qquad x\approx \begin{cases} 11.39\\ 3.28 \end{cases}

now, if we plot those two points, we get 3 regions, and then we check each of those regions to see what's the value of the first derivative.

check the 2nd picture below, that'd be the first derivative test, as you can see from the arrows, the slope increases and then decreases at 3.28, namely that critical point is the maximum, so the Volume maximizes at x = 3.28.

6 0
3 years ago
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