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3 years ago

Help on all pls i need it quick

Mathematics

1 answer:

Tpy6a [65]3 years ago

4 0

Answer:

angle 7 is congruent to angle 6 because both are vertical angles and vertical angles are congruent to one another

angle 6 and angle 10 are congruent because both are corresponding angles and corresponding angles are congruent

thus angle 7 is congruent to angle 10 because vertical angles

Step-by-step explanation:

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Find the experimental probability of rolling a 4

11Alexandr11 [23.1K]

<span />the probability of rolling a 4 = 5 times.

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3 years ago

Which of the following is the solution to the inequality below ? -6/5(4-x)<_-4(X+1/2)

LuckyWell [14K]

Answer:

The answer is the option D

$x\leq 7/13$

Step-by-step explanation:

we have

$-\frac{6}{5}(4-x)\leq -4(x+\frac{1}{2})$

Multiply by $-5$ both sides

$6(4-x)\geq 20(x+0.5)\\24-6x\geq20x+10\\24-10\geq 20x+6x\\14\geq 26x$

Rewrite

$26x\leq 14\\x\leq 14/26\\ x\leq 7/13$

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2 years ago

Name a pair of overlapping congruent triangles in each diagram. State whether the triangles are congruent by SSS, SAS, ASA, AAS,

MariettaO [177]

Answer:

ΔPTS≅ΔRTA by AAS axiom of congruency

Step-by-step explanation:

Consider ΔPQA and ΔRQS

∠PQA=∠RQS (Vertically Opposite Angles)

∠QAP=∠QSR (Complementary of two equal angles, ∠RAT and∠PST)

Due to angle sum property of a triangle, we come to the conclusion that

∠APQ=∠SRQ

Consider ΔPTS and ΔRTA

TA=TS (Given)

∠RAT=∠PST(Given)

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Therefore, ΔPTS≅ΔRTA by AAS axiom of congruency.

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3 years ago

If you run for 0.4 hours at 7 mph, how fast should you walk during the next 0.8 hours to have the average speed of 5 mph?

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Answer:

4 mph

Step-by-step explanation:

The average speed of an object is given by the total distance covered by the time taken:

$v=\frac{d}{t}$

where

d is the total distance covered

t is the time taken

in the first part, the person runs for 0.4 hours at a speed of 7 mph, so the distance covered in the 1st part is

$d_1 = v_1 t_1 = (7)(0.4)=2.8 mi$

Then the distance covered in the second part is $d_2$ , so the total distance is

$d=2.8+d_2$ (1)

The total time elapsed is 0.4 hours (first part) + 0.8 hours (second part), so

$t=0.4+0.8=1.2 h$

So we can write the average speed as

$v=\frac{2.8+d_2}{1.2}$ (1)

We want the average speed to be 5 mph,

v = 5 mph

Therefore we can rearrange eq.(1) to find d2:

$d_2 = 1.2v-2.8 = (1.2)(5)-2.8=3.2 mi$

And therefore, the speed in the second part must be

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2 years ago

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Lets say our number is 237x8

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2 years ago