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iren [92.7K]
2 years ago
12

Melissa rolls 2 fair dice and adds the results from each.

Mathematics
1 answer:
Yuri [45]2 years ago
5 0

Answer:

It would be 35/36

Step-by-step explanation:

so if you did like 1+1, 1+2, 1+3....2+1,2+2, 2+3 yada yada

the only possible number combo that would equal twelve is 6+6 and there is 36 possible combos so your answer is 35/36

- a former algebra 2 student

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Pls help me <br> What is the diagram?
olga nikolaevna [1]

Answer:

c/e = e/d

Step-by-step explanation:

These are similar triangles. When you flip the larger one clockwise to enable you to see which sides are corresponding, then you find that:

c corresponds to e'

a corresponds to b

e corresponds to d

Therefore, c/e in the original position = e'/d in the flipped position.

<u>So c/e = e/d</u>

8 0
3 years 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
64 as a mixed number
Kaylis [27]
I'm perdy sure it's 64/1 but yk it could be something else XD :D good luck!

6 0
3 years ago
Find the equation of the parabola that passes through
valentinak56 [21]

so we have the points of (0,-7),(7,-14),(-3,-19), let's plug those in the y = ax² + bx + c form, since we have three points, we'll plug each one once, thus a system of three variables, and then we'll solve it by substitution.

\bf \begin{array}{cccllll} \stackrel{\textit{point (0,-7)}}{-7=a(0)^2+b(0)+c}& \stackrel{point (7,-14)}{-14=a(7)^2+b(7)+c}& \stackrel{point (-3,-19)}{-19=a(-3)^2+b(-3)+c}\\\\ -7=c&-14=49a+7b+c&-19=9a-3b+c \end{array}

well, from the 1st  equation, we know what "c" is already, so let's just plug that in the 2nd equation and solve for "b".

\bf -14=49a+7b-7\implies -7=49a+7b\implies -7-49a=7b \\\\\\ \cfrac{-7-49a}{7}=b\implies \cfrac{-7}{7}-\cfrac{49a}{7}=b\implies -1-7a=b

well, now let's plug that "b" into our 3rd equation and solve for "a".

\bf -19=9a-3b-7\implies -12=9a-3b\implies -12=9a-3(-1-7a) \\\\\\ -12=9a+3+21a\implies -15=9a+21a\implies -15=30a \\\\\\ -\cfrac{15}{30}=a\implies \blacktriangleright -\cfrac{1}{2}=a \blacktriangleleft \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{and since we know that}}{-1-7a=b}\implies -1-7\left( -\cfrac{1}{2} \right)=b\implies -1+\cfrac{7}{2}=b\implies \blacktriangleright \cfrac{5}{2}=b \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill y=-\cfrac{1}{2}x^2+\cfrac{5}{2}x-7~\hfill

3 0
3 years ago
What fractional part of a kilometer is a meter ? ​
Crazy boy [7]

a kilometer is 1000 meters. a meter is 1 meter so the answer is:

1/1000

6 0
3 years ago
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