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3 years ago

E is the midpoint of DF, DE= 73 – 3, and EF = 4.: +3.

Mathematics

1 answer:

Damm [24]3 years ago

4 0

Answer:

i don't really understand what you are asking me

Step-by-step explanation:

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Selling price of shoes is rupees 420 including 5% GST.The original price of the shoes is-

Korolek [52]

Answer:

420

Step-by-step explanation:

selling price of shoes is rupees 420 including 5% GST.The original price of the shoes is-

7 0

3 years ago

Read 2 more answers

Pleasee helpp Write a^3 •a^4 without the exponents

Sliva [168]

Answer:

a x a x a x a x a x a x a

Step-by-step explanation:

a^3 is three a's being multiplied by each other. a^4 is four a's being multiplied by each other

5 0

3 years ago

The diagram represents a reduction of a triangle by using a scale factor of 0.8.

Alekssandra [29.7K]

Answer:

4.8 inches

Step-by-step explanation:

See comment for complete question

Represent the larger triangle with 1 and the smaller with 2.

So, we have:

$H_1 =6\ in$ -- height of 1

Required

Determine H2 --- Height of 2

To do this we apply dilation formula.

$Original * Scale\ Factor = New$

In this case:

$H_1 * Scale\ Factor = H_2$

Substitute 6 for H1 and 0.8 for Scale Factor

$6* 0.8 = H_2$

$4.8= H_2$

Hence, the height of the smaller triangle is 4.8 inches

3 0

3 years ago

Condense the expression to the logarithm of a single quantity.

vladimir1956 [14]

Answer:

$\log _3\left(\frac{x^{\frac{1}{2}}}{\left(y+8\right)^2}\right)$

Step-by-step explanation:

-> Apply log rules

$\log _3\left(x^{\frac{1}{2}}\right)-2\log _3\left(y+8\right)$

$\log _3\left(x^{\frac{1}{2}}\right)-\log _3\left(\left(y+8\right)^2\right)$

-> Apply basic math rules

$\log _3\left(\frac{x^{\frac{1}{2}}}{\left(y+8\right)^2}\right)$

7 0

2 years ago

Find the limit. (If an answer does not exist, enter DNE.) lim Δx→0 (x + Δx)2 − 3(x + Δx) + 2 − (x2 − 3x + 2) Δx

yawa3891 [41]

$\displaystyle\lim_{\Delta x\to0}\frac{(x+\Delta x)^2-3(x+\Delta x)+2-(x^2-3x+2)}{\Delta x}$

Expand the numerator as

$x^2+2x\Delta x+(\Delta x)^2-3x-3\Delta x+2-x^2+3x-2=(2x-3)\Delta x+(\Delta x)^2$

Then in the limit,

$\displaystyle\lim_{\Delta x\to0}\frac{(2x-3)\Delta x+(\Delta x)^2}{\Delta x}=\lim_{\Delta x\to0}(2x-3+\Delta x)=\boxed{2x-3}$

3 0

3 years ago