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3 years ago

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If james had seven dollars and ruby had 3,963,927 dollars. what is the sum divided by 286 times 74 subtracted then timed by 7 to

the 6th power?
$7 + 3963927 \div 286 \times 74 - {7}^{6} \times {7}^{6}$
It's my last question for my 5th grade pop quiz.

2 answers:

Nookie1986 [14]3 years ago

4 0

Answer:

Anakin im just here to talk with u!!!!!♥♥

Step-by-step explanation:

But the answer is :

Exact Form:

−179923400313 / 13

Decimal Form:

−13840261562.5385…

Mixed Number Form:

13840261562r+7

valkas [14]3 years ago

3 0

Answer:

Step-by-step explanation:

ok lets do thiiiiis

7+3963927 / 286 * 74 - 7^6 * 7^6

. first start with the exponets.

7+ 3963927 / 286 * 74 - 117649 * 117649

multiplication is next 286 * 74

now we got 7+ 3963927 / 21,164 - 13841287201 do the math and your left with

3963934 / 138141266037

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I need help thanks!!!!!

irga5000 [103]

Is this question asking for descriptions?

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3 years ago

lilavasa [31]

The count of the equilateral triangle is an illustration of areas

There are 150 small equilateral triangles in the regular hexagon

<h3>How to determine the number of equilateral triangle </h3>

The side length of the hexagon is given as:

L = 5

The area of the hexagon is calculated as:

$A = \frac{3\sqrt 3}{2}L^2$

This gives

$A = \frac{3\sqrt 3}{2}* 5^2$

$A = \frac{75\sqrt 3}{2}$

The side length of the equilateral triangle is

l = 1

The area of the triangle is calculated as:

$a = \frac{\sqrt 3}{4}l^2$

So, we have:

$a = \frac{\sqrt 3}{4}*1^2$

$a = \frac{\sqrt 3}{4}$

The number of equilateral triangles in the regular hexagon is then calculated as:

$n = \frac Aa$

This gives

$n = \frac{75\sqrt 3}{2} \div \frac{\sqrt 3}4$

So, we have:

$n = \frac{75}{2} \div \frac{1}4$

Rewrite as:

$n = \frac{75}{2} *\frac{4}1$

$n = 150$

Hence, there are 150 small equilateral triangles in the regular hexagon

Read more about areas at:

brainly.com/question/24487155

4 0

3 years ago

Determine whether the function is periodic. If it is periodic, find the period.

puteri [66]

Answer:

none of them

$f(x) = a \: \sin(b + cx) \\ t = \frac{2\pi}{ |c| } \\ t = \frac{2\pi}{ |1| } = 2\pi$

this is the right answer

6 0

3 years ago

Vector d is shown on a centimetre grid. Image attached.

nexus9112 [7]

Answer:

see explanation

Step-by-step explanation:

A column vector has the form $\left[\begin{array}{ccc}x\\y\\\end{array}\right]$

Following the direction of the arrow , the x- displacement = - 3 and y- displacement = 2 , then column vector is

$\left[\begin{array}{ccc}-3\\2\\\end{array}\right]$

h + k

= $\left[\begin{array}{ccc}4\\-3\\\end{array}\right]$ + $\left[\begin{array}{ccc}-2\\-1\\\end{array}\right]$ ← add corresponding elements from each vector

= $\left[\begin{array}{ccc}4-2\\-3-1\\\end{array}\right]$

= $\left[\begin{array}{ccc}2\\-4\\\end{array}\right]$

3 0

3 years ago

I need help with 12 13 and 14

nikdorinn [45]

Answer: Lines $\frac{}{BC}$ and $\frac{}{EF}$ are different lengths.

Step-by-step explanation:

The distance formula is $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }$ , and you can use this formula to solve for the lengths of both lines $\frac{}{BC}$ and $\frac{}{EF}$ .

For line $\frac{}{BC}$ , let $x_{1}$ = the x at point B, or 1, and let $x_{2}$ = the x at point C, or 2.

Now, let $y_{1}$ = the y at point B, or 4, and let $y_{2}$ = the y at point C, or -1.

Now, solve the formula to find the length $\frac{}{BC}$ = $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\$ .

$\frac{}{BC}$ = $\sqrt{(1-2)^{2} +(4-(-1))^2$

$\frac{}{BC}$ = $\sqrt{(-1)^{2} +(4+1)^2$

$\frac{}{BC}$ = $\sqrt{1 +5^2$

$\frac{}{BC}$ = $\sqrt{(1+25)$

$\frac{}{BC}$ = $\sqrt{26} \\$

Now, for line $\frac{}{EF}$ , let $x_{1}$ = the x at point E, or -4, and let $x_{2}$ = the x at point F, or -1.

Let $y_{1}$ = the y at point E, or -3, and let $y_{2}$ = the y at point F, or 1.

Now, solve the formula to find the length $\frac{}{EF}$ = $\sqrt{(x_1-x_2) ^{2}+(y_1-y_2) ^{2} }\\$ .

$\frac{}{EF}$ = $\sqrt{(-4-(-1))^{2} +(-3-1)^2$

$\frac{}{EF}$ = $\sqrt{(-4+1)^{2} +(-4)^2$

$\frac{}{EF}$ = $\sqrt{(-3)^2+16$

$\frac{}{EF}$ = $\sqrt{(9+16)$

$\frac{}{EF}$ = $\sqrt{25}$

$\frac{}{EF}$ = 5

Now, look back at $\frac{}{BC}$ . The two lines have different lengths, so you have now justified the fact that they are not the same.

Questions 13 and 14 would be solved in much the same way- but please let me know if you want me to show the work for those as well!

7 0

3 years ago