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3 years ago

10

Sandra is asked to report on the recent increase in popularity of an application based on the number

1 answer:

aalyn [17]3 years ago

3 0

Answer:

Step-by-step explanation:

Report the number of downloads per day. By determining how this number is changing / increasing it will be possible to spot and describe an upward trend if there is one.

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Describe a sequence of transformations that can be performed to polygon ABCDE to prove that it is similar to polygon FGHIJ

maxonik [38]

Answer:

answer below

Step-by-step explanation:

ABCDE go through dilation over center (6 , -2) with factor of 1/2 to FGHIJ

AB // FG slope: -2 , √20:√5 = 2: 1

BC // GH // X axis 8:4 = 2:1

CD // HI, slope= 1 , √8:√2 = 2:1

DE // IJ // x axis, 4:2 = 2:1

EA // JF // y axis, 2:1

8 0

2 years ago

Ganezh [65]

Answer:

$x^{2} -9x$

Step-by-step explanation:

I will upload a picture with my work.

4 0

3 years ago

PLEASE HELP (100 POINTS)

omeli [17]

$\\ \rm\Rrightarrow 2sin^2x=1-cosx$

$\\ \rm\Rrightarrow 2sin^2x=1-1+sin^2x$

$\\ \rm\Rrightarrow 2sin^2x-sin^2x=0$

$\\ \rm\Rrightarrow sin^2x=0$

$\\ \rm\Rrightarrow sinx=0$

$\\ \rm\Rrightarrow x\in n\pi$

6 0

2 years ago

Hoping someone will help me out

pogonyaev

Answer:

It does not converge

Step-by-step explanation:

7 0

3 years ago

Read 2 more answers

Rex, Paulo, and Ben are standing on shore watching for dolphins. Paulo sees one surface directly in front of him about a hundred

Ksivusya [100]

1. $m\angle BAC=m\angle CAD,\ m\angle ACB=m\angle ADC=90^{\circ}$ , then $m\angle ABC=m\angle ACD$ and triangles ADC and ACB are similar by AAA theorem.

2. The ratio of the corresponding sides of similar triangles is constant, so

$\dfrac{AC}{AB}= \dfrac{AD}{AC}$ .

3. Knowing lengths you could state that $\dfrac{b}{c}= \dfrac{e}{b}$ .

4. This ratio is equivalent to $b^2=ce$ .

5. $m\angle ABC=m\angle CBD,\ m\angle ACB=m\angle CDB=90^{\circ}$ , then $m\angle BAC=m\angle BCD$ and triangles BDC and BCA are similar by AAA theorem.

6. The ratio of the corresponding sides of similar triangles is constant, so

$\dfrac{BC}{BD}= \dfrac{AB}{BC}$ .

7. Knowing lengths you could state that $\dfrac{a}{d}= \dfrac{c}{a}$ .

8. This ratio is equivalent to $a^2=cd$ .

9. Now add results of parts 4 and 8:

$b^2+a^2=ce+cd$ .

10. c is common factor, then:

$b^2+a^2=c(e+d)$ .

11. Since $e+d=c$ you have $a^2+b^2=c\cdot c=c^2$ .

7 0

3 years ago

Read 2 more answers