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d1i1m1o1n [39]
3 years ago
8

Molly earns $6 for each hour she babysits plus $10 to cover any personal

Mathematics
2 answers:
qwelly [4]3 years ago
8 0

Answer:

44

Step-by-step explanation:

SIZIF [17.4K]3 years ago
5 0

Answer:

The answer is C or 52!!

Step-by-step explanation:

Brainliest Please!

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How far is the little guy from the flagpole (assuming a right angle between the flagpole and the ground)? Explain how you got yo
Naya [18.7K]

Answer:

b = 9 ft

Step-by-step explanation:

Pythagorean Theorem:

a^2 + b ^2 = c^2

Given:

a = 12

b = ?

c = 15

Plug in values:

12^2 + b^2 = 12^2

Solve for b^2:

b^2 = 15^2 - 12^2

b^2 = 225 - 144

b^2 = 81

Take square root of both sides to find b:

sqrt(b^2) = sqrt(81)

b = 9


Hope this helps! :)

7 0
3 years ago
Given: AE ≅ CE ; DE ≅ BE Prove: ABCD is a parallelogram. Parallelogram A B C D is shown. Diagonals are drawn from point A to poi
nikklg [1K]

Answer:

The proof is below

Step-by-step explanation:

Given a parallelogram ABCD. Diagonals AC and BD intersect at E. We have to prove that AE is congruent to CE and BE is congruent to DE i.e diagonals of a parallelogram bisect each other.

In ΔACD and ΔBEC

AD=BC              (∵Opposite sides of a parallelogram are equal)

∠DAC=∠BCE       (∵Alternate angles)

∠ADC=∠CBE        (∵Alternate angles)

By ASA rule, ΔACD≅ΔBEC

By CPCT(Corresponding Parts of Congruent triangles)

AE=EC and DE=EB

Hence, AE is conruent to CE and BE is congruent to DE

6 0
3 years ago
A teacher surveyed her students about the amount of time the students
LiRa [457]

Answer:

c

Step-by-step explanation:

I just had the exact same question on a worksheet

5 0
3 years ago
Which equation has exactly one solution?
KIM [24]
<h3>Answer: Choice B</h3>

===================================================

Explanation:

Let's look at choice A

The left hand side simplifies as follows

\frac{2+6x}{2}\\\\\frac{2}{2}+\frac{6x}{2}\\\\1+3x\\\\

This means the equation \frac{2+6x}{2}=3x+1\\\\ is the same as 1+3x=3x+1\\\\. Notice how the left and right sides are identical expressions. They both involve 3x and a 1. This means this equation has infinitely many solutions. We can rule out choice A.

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

I'll skip choice B and come back

Let's move onto choice C.

The given equation \frac{15x-5}{3}=5x+4\\\\ transforms into 5x-\frac{5}{3}=5x+4\\\\ through similar steps as choice A shows.

If we were to subtract 5x from both sides, all of the x terms go away on both sides. We'd be left with the equation -\frac{5}{3}=4\\\\ which is always false, which means that the original equation is always false.

Therefore, the equation of choice C has no solutions. We consider this equation inconsistent. We rule out choice C because of all this.

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

Choice D is pretty much the same as choice A even down to the conclusion of getting infinitely many solutions. The left hand side simplifies fully into the right hand side. The two sides are identical; hence, we have an identity. An identity is true for all allowed x values in the domain.

In other words, whatever x value you pick it will make the equation true. So we can rule this out because we want one solution only.

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

Only choice B is left. You'll notice that after simplifying the left hand side, the x term on the left side doesn't turn into 2x. So there's no way to get the options of "infinitely many solutions" or "no solutions". Those two cases only happen when we have the same x terms on both sides.

Put another way, the two slopes of each side are different which produces nonparallel lines that intersect at exactly one location. This intersection location is the solution. Specifically, the x coordinate of the (x,y) location is the solution.

5 0
3 years ago
Using the U- Substitution u=sqrt(2x), integral form 2-8 dx/ sqrt(2x) + 1 is equivalent to ...
AleksAgata [21]
We will use u-substitute:u= \sqrt{2x} , \frac{du}{dx}= \frac{1}{ \sqrt{2x} }= \frac{1}{u}Then for substitution:dx=u du. and integral becomes:\int { \frac{u}{u+1} } \, du = \int { \frac{u+1-1}{u+1} } \, du= \int{1} \, du- \int { \frac{1}{u-1} } \, dx=u-ln(u+1)=\sqrt{2x}-ln( \sqrt{2x}+1). Now we will change the values of limits: \sqrt{16}-ln( \sqrt{16}+1)-( \sqrt{4}-ln( \sqrt{4}+1))=4-ln(5)-2+ln(3)=2+ln(0,6)=2-0.51=1.49

8 0
3 years ago
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