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sergiy2304 [10]

3 years ago

15

Dylan works 30 hours a week. His hours are decreased by 20%. How many hours does he now work each week? 15 20 24 28

1 answer:

fredd [130]3 years ago

3 0

Dylan now must work 24 hours a week.

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Can someone explain to me better on identifying linear functions? Like how would I know the difference when identifying them? Ar

Margaret [11]

Answer:

A linear function is a straight line and a non-linear function is not a straight line.

Step-by-step explanation:

7 0

2 years ago

I need help I don’t understand

Arte-miy333 [17]

Use mathpapa.com or cymath.com

7 0

3 years ago

Can someone please help me solve these two problems and explain them to me? About the one with proportions though... I've always

NeTakaya

Answers:

A) $-\frac{11}{4}$

C) $m_{1}=\frac{Fr^{2}}{G m_{2}}$

Step-by-step explanation:

<u>Part 1:</u>

We have the followig equation:

$\frac{x-1}{3}-\frac{x+4}{5}=\frac{4x-1}{8}$

Calculating the least common multiple (l.c.m) in the denominator in the left side of the equation, being l.c.m=15:

$\frac{5(x-1)-3(x+4)}{15}=\frac{4x-1}{8}$

Solving for the left part of the equation:

$\frac{2x-17}{15}=\frac{4x-1}{8}$

Operating with cross product:

$8(2x-17)=15(4x-1)$

Applying the distributive property:

$16x-136=60x-15$

Isolating $x$ :

$x=-\frac{121}{44}$

Dividing numerator and denominator by 11:

$x=-\frac{11}{4}$ Hence, the correct option is A

<u>Part 2:</u>

We have the followig equation:

$F=\frac{G m_{1} m_{2}}{r^{2}}$

Operating with cross product:

$Fr^{2}=G m_{1} m_{2}$

Isolating $m_{1}$ :

$m_{1}=\frac{Fr^{2}}{G m_{2}}$ Hence, the correct option is C

5 0

3 years ago

Wilma bought a skateboard for 20% off. The normal price was 120.90. How much did Wilma pay for the skateboard, not including tax

evablogger [386]

The answer is 96.72

7 0

3 years ago

100 POINTS!! OMG YOU NEED TO SOLVE THIS!

natita [175]

Answer:

The correct answer is: $\Delta BDE \cong \Delta BFK$ by <em>rule</em> ASA rule of congruence.

Step-by-step explanation:

First let us prove $\Delta BDE \cong \Delta BFK$ by rule ASA (rule of congruence).

<u>Congruent side:</u>

$\overline{BD} \cong \overline{BF}$ (Given)

<u>Congruent angles:</u>

1. By definition of perpendicular,

$\angle{BFK} = 90 \textdegree \ (Since \ \overline{FK} \ is \ perpendicular \ to \ \overline{AB} \ (\overline{FK} \perp \overline{AB} =Given))$

Also,

$\angle{BDE} = 90 \textdegree \ (Given)$

Therefore,

$\angle{BDE} \cong \angle{BFK}$

or you can say,

$\angle{D} \cong \angle{F}$

2. Common angle between $\Delta BDE \ and \ \Delta BFK$ is $\angle{B}$

In a nutshell, in $\Delta BDE$ ,

$\angle{B}$ (Angle)

$\overline{BD}$ (Side)

$\angle{BDE}$ (Angle)

are congruent to the following angle, side and angle of $\Delta BFK$ :

$\angle{B}$ (Angle)

$\overline{BF}$ (Side)

$\angle{BFK}$ (Angle)

Therefore, by ASA rule of congruence, we can say $\Delta BDE \cong \Delta BFK$ .

<em>Since both triangles are congruent</em>, the sides $\overline{ED}$ and $\overline{FK}$ are also congruent.

5 0

3 years ago

Read 2 more answers