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Andrews [41]
3 years ago
6

Which expression is equivalent to the expression 3^7x3^-4

Mathematics
1 answer:
kykrilka [37]3 years ago
5 0

Answer:

3^7 * 3^(-4) = 3^(7 - 4) = 3^3 = 27

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Sisqo score 10 points in the first half of the game. In the second half he made four shots, all three pointers. Which of the fol
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Answer:

You didn't provided the answer choices, but the answer is he scored 22 points in total.

Step-by-step explanation:

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What is a correct congruence statement for the triangles shown? Enter your answer in the box. ​ △QDJ≅△ ​ A triangle Q D J. The b
almond37 [142]

Answer: The correct congruence statement is \triangle QDJ\cong \triangle MCW.

Explanation:

It is given that A triangle Q D J. The base D J is horizontal and side Q D is vertical. Another triangle M C W is made. The base C W and side M C are neither horizontal nor vertical. Triangle M C W is to the right of triangle Q D J.

The sides Q D and M C are labeled with a single tick mark. The sides D J and C W are labeled with a double tick mark. The sides Q J and M W are labeled with a triple tick mark.

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From the figure it is noticed that

QD\cong MC

DJ\cong CW

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So by SSS rule of congruence we can say that \triangle QDJ\cong \triangle MCW.

4 0
3 years ago
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19. answer this question.​
Irina-Kira [14]
<h3>♫ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ♫</h3>

➷ final = original x multiplier^n

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Substitute the values in:

final = 3800 x 1.054^3

final = 4449.440763

The closest answer is A

<h3><u>✽</u></h3>

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Look at the street map. What street intersects with Polk Street?
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Geary street :D

they intersect here

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2 years ago
a crowbar of 2cm is used to lift an object if 800N. if the effort arm is 260cm. calculate the effort applied.​
Gelneren [198K]

Answer:

solution:Here

Given,

Load=800 N

Load distance=2cm

Effort distance=260cm

Effort applied=?

We know that,

Load × Load distance = Effort × Effort distance

or, 800×2 = e × 260

or,1600 = 260e

or,e = 1600/260

Thus, e = <u>6.15</u><u> </u><u>N</u>

<u>Hence</u><u>,</u><u> </u><u>The</u><u> </u><u>effort</u><u> </u><u>applied</u><u> </u><u>=</u><u> </u><u>6.15</u><u> </u><u>N</u>

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