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Vsevolod [243]
3 years ago
15

A candle is 7 in. tall after burning for 1 h. The same candle is in. tall after burning for 4 h. How tall will the candle be aft

er burning for 6 h?
Mathematics
1 answer:
V125BC [204]3 years ago
8 0

After burning 6 hours the candle will be 4 1/2 in. tall.

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The answer is ten. Perimeter is all the sides added together. Area is length times width.

Step-by-step explanation:

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3 years ago
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Solve the equation <br><br> 14 = 9 - p
morpeh [17]

14 = 9 - p

Move 9 to the other side

Sign changes from +9 to -9

14-9= 9-9-p

14-9= -p

5=-p

Mutiply both sides by -1 to get + p

(5)(-1)= (-p)(-1)

p= -5

Answer: p= -5

3 0
3 years ago
B. Find the equation of the line through (-3, 3) with slope 4.
yarga [219]

Answer:

Step-by-step explanation:

use the formula y=mx+b. the first number in the parenthesis is -3 so that is x. 3 is y and the slope is m

so it is 3=4(-3)+b

4 0
3 years ago
What is the inverse and restricted domain of the equation 8x^2-3
kari74 [83]

Answer:  \bold{y=\pm \dfrac{\sqrt{2(x+3)}}{4},\qquad x\neq -3}

<u>Step-by-step explanation:</u>

y = 8x² - 3            (Restriction: none -  x is All Real Numbers)

The inverse is when you swap the x's and y's and then solve for y

x = 8y² - 3            <em>swapped the x and y</em>

x + 3 = 8y²           <em>added 3 to both sides</em>

\dfrac{x+3}{8}=y^2           <em>divided both sides by 8</em>

\sqrt{\dfrac{x+3}{8}}=\sqrt{y^2}           <em>square rooted both sides</em>

\pm \sqrt{\dfrac{x+3}{8}}=y           <em>simplified</em>

\pm \sqrt{\dfrac{x+3}{8}\bigg(\dfrac{2}{2}\bigg)}=y           rationalized the denominator

\pm \sqrt{\dfrac{2(x+3)}{16}}=y           <em>simplified</em>

\pm \dfrac{\sqrt{2(x+3)}}{4}=y           <em>simplified</em>

<u>Restriction:</u>

The radical <em>(inside the square root sign)</em> cannot be negative

→  2(x + 3) ≥ 0

      x + 3 ≥ 0         <em>divided both sides by 2</em>

      x       ≥ -3         <em>subtracted 3 from both sides</em>



4 0
3 years ago
Help please!!! Thank you
Elena L [17]

<u>Given</u>:

The given expression is \frac{4 d+28}{12 d+96} \cdot \frac{d^{2}+14 d+48}{d^{2}+9 d+14}

We need to multiply the terms.

<u>Multiplication of the terms:</u>

Before multiplying the terms, first we shall find the factors of the quadratic equations.

Thus, we have;

\frac{4 d+28}{12 d+96} \cdot \frac{(d+6)(d+8)}{(d+2)(d+7)}

Factor out the common terms, we get;

\frac{4 (d+7)}{12 (d+8)} \cdot \frac{(d+6)(d+8)}{(d+2)(d+7)}

Let us cancel the common terms from the above expression.

Thus, we have;

\frac{4 }{12 } \cdot \frac{(d+6)}{(d+2)}

Simplifying, the term, we get;

\frac{1}{3 } \cdot \frac{(d+6)}{(d+2)}

Now, we shall multiply the terms.

Hence, multiplying the terms, we get;

\frac{(d+6)}{3(d+2)}

Thus, the multiplied value of the given expression is \frac{(d+6)}{3(d+2)}

6 0
3 years ago
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