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anygoal [31]
2 years ago
11

Help please and thanks! I always give brainliest!

Mathematics
1 answer:
Alika [10]2 years ago
3 0

Answer:

X = 18

y = 9

Step-by-step explanation:

Basically the 2 diagnols bisect each other making them congruent. so that makes

x-3 = 15 and y + 3 = 12

simplify and

the. you simplify and get x = 18 and y =9

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Can someone please help me
Eva8 [605]

Answer:

B and F should be right since the angle in the middle of them anchors in the the middle of the circle

8 0
3 years ago
The difference of two is 4. The larger is 8 less than twice the smaller. What are the two numbers ?
siniylev [52]

Answer:

{12, 16}

Step-by-step explanation:

The difference of two <em>numbers (x and y)</em> is 4.  Thus, if y is the larger, then y - x = 4.  Also, y = 2x - 8.

So we need to solve the system of linear equations

y - x = 4

y = 2x - 8

Subbing 2x - 8 for y, in the first equation, we get  (2x - 8) - x = 4, or

x - 8 = 4, and conclude that x = 12.  Then, because y - x = 4, y must be 16.

The numbers are 12 and 16.

 

5 0
3 years ago
20 pts please get this right
Serga [27]

Answer:

The answer is C. Citizens vote for the president. Hope this helps! Have a great day!

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
PLZ HELP!!! Use limits to evaluate the integral.
Marrrta [24]

Split up the interval [0, 2] into <em>n</em> equally spaced subintervals:

\left[0,\dfrac2n\right],\left[\dfrac2n,\dfrac4n\right],\left[\dfrac4n,\dfrac6n\right],\ldots,\left[\dfrac{2(n-1)}n,2\right]

Let's use the right endpoints as our sampling points; they are given by the arithmetic sequence,

r_i=\dfrac{2i}n

where 1\le i\le n. Each interval has length \Delta x_i=\frac{2-0}n=\frac2n.

At these sampling points, the function takes on values of

f(r_i)=7{r_i}^3=7\left(\dfrac{2i}n\right)^3=\dfrac{56i^3}{n^3}

We approximate the integral with the Riemann sum:

\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{112}n\sum_{i=1}^ni^3

Recall that

\displaystyle\sum_{i=1}^ni^3=\frac{n^2(n+1)^2}4

so that the sum reduces to

\displaystyle\sum_{i=1}^nf(r_i)\Delta x_i=\frac{28n^2(n+1)^2}{n^4}

Take the limit as <em>n</em> approaches infinity, and the Riemann sum converges to the value of the integral:

\displaystyle\int_0^27x^3\,\mathrm dx=\lim_{n\to\infty}\frac{28n^2(n+1)^2}{n^4}=\boxed{28}

Just to check:

\displaystyle\int_0^27x^3\,\mathrm dx=\frac{7x^4}4\bigg|_0^2=\frac{7\cdot2^4}4=28

4 0
3 years ago
Plsss helpppp what’s the shape?
Alborosie
The shape shown here is a triangular pyramid, because it has a triangular base and a pointy tip.
4 0
3 years ago
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