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2 years ago

Aisha 3 times as many chocolates as Abdul. Together they have 24 chocolates, How many chocolates does each child have?

Mathematics

1 answer:

ch4aika [34]2 years ago

5 0

Answer:

3x+X =24

4x=24

X=24/4

X=6

Aisha= 3x6

=18

Abdulla=1x6

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The price of two pairs and six apples is $14. The price of three pairs and nine apples is $21. Can you determine the unit prices

deff fn [24]

Answer:

Step-by-step explanation:

The second purchase is exactly half-again larger than the first purchase, so sheds no light on the relative costs of the items. The per-item costs can be found when the ratio of items purchased is different from one buy to another.

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We can only say that the cost of 1 pear and 3 apples is $7.

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3 years ago

What is the completely factored form of d4 -8d2 + 16

irinina [24]

According to Vieta's Formulas, if $x_1,x_2$ are solutions of a given quadratic equation:

$ax^2+bx+c=0$

Then:

$a(x-x_1)(x-x_2)$ is the completely factored form of $ax^2+bx+c$ .

If choose $x=d^2$ , then:

$\displaystyle x^2-8x+16=0\\\\x_{1,2}= \frac{8\pm \sqrt{64-64} }{2}=4$

So, according to Vieta's formula, we can get:

$x^2-8x+16=(x-4)(x-4)= (x-4)^2$

But $x=d^2$ :

$d^4-8d^2+16=(d^2-4)^2=[(d+2)(d-2)]^2=(d+2)^2(d-2)^2$

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2 years ago

What two-dimensional figure will be created by a<br> horizontal cross section?

Nuetrik [128]

Answer: First answer: hexagon

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Step-by-step explanation: right on edge 2020

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2 years ago

Read 2 more answers

If four times the sum of a number and 3 is divided by 2, the quotient is 0: find the number

AveGali [126]

The other number is -3.

Step-by-step explanation:

Step 1; First we develop formulae for the given information. A number is added with 3 and the resulting sum is multiplied by 4. This number is then divided by 2 and the quotient is 0. Assume the unknown number is x, then the given information is as follows

The quotient of $\frac{4(x+3)}{2}$ is 0. So $\frac{4(x+3)}{2}$ = 0.

Step 2; Now we solve the above equation. The denominator is taken to the RHS and the RHS remains zero.

$\frac{4(x+3)}{2}$ = 0, 4(x+3)= 0, 4x+ 12 = 0, 4x = -12, x = $\frac{-12}{4}$ = -3.

So the other number is -3.

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3 years ago

Can I get help with finding the Fourier cosine series of F(x) = x - x^2

trapecia [35]

Assuming you want the cosine series expansion over an arbitrary symmetric interval $[-L,L]$ , $L\neq0$ , the cosine series is given by

$f_C(x)=\dfrac{a_0}2+\displaystyle\sum_{n\ge1}a_n\cos nx$

You have

$a_0=\displaystyle\frac1L\int_{-L}^Lf(x)\,\mathrm dx$
$a_0=\dfrac1L\left(\dfrac{x^2}2-\dfrac{x^3}3\right)\bigg|_{x=-L}^{x=L}$
$a_0=\dfrac1L\left(\left(\dfrac{L^2}2-\dfrac{L^3}3\right)-\left(\dfrac{(-L)^2}2-\dfrac{(-L)^3}3\right)\right)$
$a_0=-\dfrac{2L^2}3$

$a_n=\displaystyle\frac1L\int_{-L}^Lf(x)\cos nx\,\mathrm dx$

Two successive rounds of integration by parts (I leave the details to you) gives an antiderivative of

$\displaystyle\int(x-x^2)\cos nx\,\mathrm dx=\frac{(1-2x)\cos nx}{n^2}-\dfrac{(2+n^2x-n^2x^2)\sin nx}{n^3}$

and so

$a_n=-\dfrac{4L\cos nL}{n^2}+\dfrac{(4-2n^2L^2)\sin nL}{n^3}$

So the cosine series for $f(x)$ periodic over an interval $[-L,L]$ is

$f_C(x)=-\dfrac{L^2}3+\displaystyle\sum_{n\ge1}\left(-\dfrac{4L\cos nL}{n^2L}+\dfrac{(4-2n^2L^2)\sin nL}{n^3L}\right)\cos nx$

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3 years ago