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3 years ago

a pound of chocolate costs 8 dollars. Felipe buys p pounds. write an equation to represent the total cost c that felipe pays

Mathematics

1 answer:

alisha [4.7K]3 years ago

6 0

Answer:

c=8p

Step-by-step explanation:

So, we know that 1 pound of chocolate will cost $8. Felipe is going to buy "p" pounds of chocolate, and "p" means "any number"; so Felipe is going to buy any number of pounds of chocolate.

The formula we would use to calculate this is cost of one pound of chocolate multiplied by number of pounds of chocolate. And the way we can form this algebraically is by writing c=8p.

The "c" stands for total cost and we don't have to put the multiply sign when using algebra, as long as the figures are touching.

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Helppp guys i really don't get it

AysviL [449]

9514 1404 393

Answer:

(b) ... confirm ∠C≅∠E

Step-by-step explanation:

The first step is to read and understand the problem statement. It is asking for a way to prove ΔABC ~ ΔADE by AA similarity. That means you want to show any two of the three ...

∠A≅∠A
∠B≅∠D
∠C≅∠E

For this purpose, lengths of line segments are irrelevant (eliminating the last two answer choices). The fact that ∠A≅∠A is given, so you only need to find an answer choice that will show one of the last two angle congruences.

Obviously, showing ∠B≅∠E (choice A) is not relevant to the problem.

The only answer choice that is relevant to the question is the second one, showing ∠C≅∠E.

8 0

3 years ago

Please help me with this problem I tryird but nothing has came up so please help me thank you so much

LenaWriter [7]

To compare fractions , you need to have same denominator.

Just multiply the Numerator and denominator of the fraction by required number so as to get the common denominator.

for example, In this case , the common denominator is 12.
the first fraction is converted as shown, 5/3=5*4/3*4=20/12
similarly, you can convert others.
Once you have common denominator, just compare numerators.

4 0

4 years ago

What is 5 1/2% simplified as a fraction in simplistic form

zzz [600]

Answer:

11/200

Step-by-step explanation:

=5.5%

=5.5/100

=55/1000

=55/5

------------

1000/5

= 11/200

8 0

3 years ago

True of false: If a right triangle has legs of length 8 and 15, then we know the length of the hypotenuse is the square root of

nasty-shy [4]

The answer is true, it will equal to 17!

7 0

3 years ago

Read 2 more answers

Solve for x in the equation 2x^2+3x-7=x^2+5x+39

Shalnov [3]

Hey there, hope I can help!

$\mathrm{Subtract\:}x^2+5x+39\mathrm{\:from\:both\:sides}$
$2x^2+3x-7-\left(x^2+5x+39\right)=x^2+5x+39-\left(x^2+5x+39\right)$

Assuming you know how to simplify this, I will not show the steps but can add them later on upon request
$x^2-2x-46=0$

Lets use the quadratic formula now
$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$
$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:} a=1,\:b=-2,\:c=-46: x_{1,\:2}=\frac{-\left(-2\right)\pm \sqrt{\left(-2\right)^2-4\cdot \:1\left(-46\right)}}{2\cdot \:1}$

$\frac{-\left(-2\right)+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

Multiply the numbers 2 * 1 = 2
$\frac{2+\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2+\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ \sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}$

$\mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \sqrt{\left(-2\right)^2+1\cdot \:4\cdot \:46} \ \textgreater \ \left(-2\right)^2=2^2, 2^2 = 4$

$\mathrm{Multiply\:the\:numbers:}\:4\cdot \:1\cdot \:46=184 \ \textgreater \ \sqrt{4+184} \ \textgreater \ \sqrt{188} \ \textgreater \ 2 + \sqrt{188}$
$\frac{2+\sqrt{188}}{2} \ \textgreater \ Prime\;factorize\;188 \ \textgreater \ 2^2\cdot \:47 \ \textgreater \ \sqrt{2^2\cdot \:47}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{ab}=\sqrt[n]{a}\sqrt[n]{b} \ \textgreater \ \sqrt{47}\sqrt{2^2}$

$\mathrm{Apply\:radical\:rule}: \sqrt[n]{a^n}=a \ \textgreater \ \sqrt{2^2}=2 \ \textgreater \ 2\sqrt{47} \ \textgreater \ \frac{2+2\sqrt{47}}{2}$

$Factor\;2+2\sqrt{47} \ \textgreater \ Rewrite\;as\;1\cdot \:2+2\sqrt{47}$
$\mathrm{Factor\:out\:common\:term\:}2 \ \textgreater \ 2\left(1+\sqrt{47}\right) \ \textgreater \ \frac{2\left(1+\sqrt{47}\right)}{2}$

$\mathrm{Divide\:the\:numbers:}\:\frac{2}{2}=1 \ \textgreater \ 1+\sqrt{47}$

Moving on, I will do the second part excluding the extra details that I had shown previously as from the first portion of the quadratic you can easily see what to do for the second part.

$\frac{-\left(-2\right)-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1} \ \textgreater \ \mathrm{Apply\:rule}\:-\left(-a\right)=a \ \textgreater \ \frac{2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)}}{2\cdot \:1}$

$\frac{2-\sqrt{\left(-2\right)^2-\left(-46\right)\cdot \:1\cdot \:4}}{2}$

$2-\sqrt{\left(-2\right)^2-4\cdot \:1\cdot \left(-46\right)} \ \textgreater \ 2-\sqrt{188} \ \textgreater \ \frac{2-\sqrt{188}}{2}$

$\sqrt{188} = 2\sqrt{47} \ \textgreater \ \frac{2-2\sqrt{47}}{2}$

$2-2\sqrt{47} \ \textgreater \ 2\left(1-\sqrt{47}\right) \ \textgreater \ \frac{2\left(1-\sqrt{47}\right)}{2} \ \textgreater \ 1-\sqrt{47}$

Therefore our final solutions are
$x=1+\sqrt{47},\:x=1-\sqrt{47}$

Hope this helps!

8 0

3 years ago

Read 2 more answers