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Alla [95]
3 years ago
6

i am a 2 digit number. I am the difference of two numbers whose sum is 36. The larger of these numbers is twice the smaller. Wha

t number am I?
Mathematics
1 answer:
inn [45]3 years ago
7 0
I am the diffrence of the two numbers... sum is 36 so the larger is double the size of the smaller the only numbers that work are 24 and 12 so the diffrence would be 12
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Andrea and sula ran a race on a track that was 100 meters long.When andrea crossed the finish line sula was exactly 5 meters beh
cestrela7 [59]
Andrea will be at 95 meters and Sula would win I think.
7 0
3 years ago
What is the value of x?<br> 4x − 6 = 18
tensa zangetsu [6.8K]
You need to get rid of -6 by adding 6, you also need to add 6 to 18.

4x = 24

Now you need to get rid of the 4 attached to the variable, you do this by dividing 4 since it is multiplying, also dividing 4 and 24

x = 6

So the value of x is 6

Hope this helps! :)
8 0
3 years ago
Read 2 more answers
Karin wants to use the distributive property to mentally find the value of 19 x 42+ 19x 58. Which expression can she use?
olasank [31]

Answer:

19(42 + 58)

Step-by-step explanation:

Assuming x = multiply, you are trying to solve the distributive property. Divide common factors from both terms (in this case 19).

~

7 0
3 years ago
Read 2 more answers
What is the approximate volume of a cylinder with a diameter of 12 meters and a height of 7 meters? Use 3.14 for pi.
kkurt [141]
V=\pi r^2 h
r - radius, h - height
The diameter is twice the radius.

d=12 \ m \\&#10;r=\frac{12}{2} \ m = 6 \ m \\&#10;h=7 \ m \\ \\&#10;V=\pi \times 6^2 \times 7=\pi \times 36 \times 7=252 \pi \approx 252 \times 3.14=791.28 \approx 791.3

The answer is c. 791.3 m³.
3 0
4 years ago
Which of the equations below could be the equation of this parabola?
nirvana33 [79]

Answer:

 y=-4x^2  is the equation of this parabola.

Step-by-step explanation:

Let us consider the equation

y=-4x^2

\mathrm{Domain\:of\:}\:-4x^2\::\quad \begin{bmatrix}\mathrm{Solution:}\:&\:-\infty \:

\mathrm{Range\:of\:}-4x^2:\quad \begin{bmatrix}\mathrm{Solution:}\:&\:f\left(x\right)\le \:0\:\\ \:\mathrm{Interval\:Notation:}&\:(-\infty \:,\:0]\end{bmatrix}

\mathrm{Axis\:interception\:points\:of}\:-4x^2:\quad \mathrm{X\:Intercepts}:\:\left(0,\:0\right),\:\mathrm{Y\:Intercepts}:\:\left(0,\:0\right)

As

\mathrm{The\:vertex\:of\:an\:up-down\:facing\:parabola\:of\:the\:form}\:y=a\left(x-m\right)\left(x-n\right)

\mathrm{is\:the\:average\:of\:the\:zeros}\:x_v=\frac{m+n}{2}

y=-4x^2

\mathrm{The\:parabola\:params\:are:}

a=-4,\:m=0,\:n=0

x_v=\frac{m+n}{2}

x_v=\frac{0+0}{2}

x_v=0

\mathrm{Plug\:in}\:\:x_v=0\:\mathrm{to\:find\:the}\:y_v\:\mathrm{value}

y_v=-4\cdot \:0^2

y_v=0

Therefore, the parabola vertex is

\left(0,\:0\right)

\mathrm{If}\:a

\mathrm{If}\:a>0,\:\mathrm{then\:the\:vertex\:is\:a\:minimum\:value}

a=-4

\mathrm{Maximum}\space\left(0,\:0\right)

so,

\mathrm{Vertex\:of}\:-4x^2:\quad \mathrm{Maximum}\space\left(0,\:0\right)

Therefore,  y=-4x^2  is the equation of this parabola. The graph is also attached.

7 0
3 years ago
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