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Marrrta [24]
3 years ago
8

Can someone please help me with this

Mathematics
1 answer:
Lynna [10]3 years ago
7 0

A is the answer to your question

please rate excellent need it for my rankup


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Enter the lengths of QS and TV<br><br>QS=<br>TV=​
kicyunya [14]
3v + 2 = 7v
2 = 4v
1/2 = v

QS = 3v + 2
= 3(1/2)+ 2
= 1.5 + 2
= 3.5

TV = 7v
= 7(1/2)
= 3.5
5 0
2 years ago
When graphing the inequality y ≤ 2x − 4, the boundary line needs to be graphed first. Which graph correctly shows the boundary l
Volgvan

Answer:

See Below

Step-by-step explanation:

The boundary line follows the graph of y = 2x - 4 because it represents the line of maximum or minimum values the graph could take. y = 2x - 4 has a y-intercept of -4 because it is in slope-intercept form. It also has a slope of 2, so to find another point on the line, you can go up two and right one. Also, this inequality has a solid line because it is "less than or equal to" and a solid line represents the values that equal.

6 0
2 years ago
If a photo measuring 4 inches by 6 inches is placed in a photo frame that is 2 inches wide all around, what percent of the frame
BARSIC [14]
It would be 2 by 2 in inches as the frame can only fit that much , which the area framed would be 4 inches as you multiply 2×2=4
8 0
3 years ago
Read 2 more answers
If 28% of students in College are near-sighted, the probability that in a randomly chosen group of 20 College students, exactly
cluponka [151]

Answer: (D) 16%

Step-by-step explanation:

Binomial probability formula :-

P(x)=^nC_xp^x(1-p)^n-x, where n is the sample size , p is population proportion and P(x) is the probability of getting success in x trial.

Given : The proportion of students in College are near-sighted : p= 0.28

Sample size : n= 20

Then, the the probability that in a randomly chosen group of 20 College students, exactly 4 are near-sighted is given by :_

P(x=4)=^{20}C_4(0.28)^4(1-0.28)^{20-4}\\\\=\dfrac{20!}{4!16!}(0.28)^4(0.72)^{16}\\\\=0.155326604912\approx0.16\%

Hence, the probability that in a randomly chosen group of 20 College students, exactly 4 are near-sighted is closest to 16%.

7 0
3 years ago
In ΔABC, m∠B = m∠C. The angle bisector of ∠B meets AC at point H and the angle bisector of ∠C meets AB at point K. Prove that BH
solniwko [45]

Answer:

See explanation

Step-by-step explanation:

In ΔABC, m∠B = m∠C.

BH is angle B bisector, then by definition of angle bisector

∠CBH ≅ ∠HBK

m∠CBH = m∠HBK = 1/2m∠B

CK is angle C bisector, then by definition of angle bisector

∠BCK ≅ ∠KCH

m∠BCK = m∠KCH = 1/2m∠C

Since m∠B = m∠C, then

m∠CBH = m∠HBK = 1/2m∠B = 1/2m∠C = m∠BCK = m∠KCH   (*)

Consider triangles CBH and BCK. In these triangles,

  • ∠CBH ≅ ∠BCK (from equality (*));
  • ∠HCB ≅ ∠KBC, because m∠B = m∠C;
  • BC ≅CB by reflexive property.

So, triangles CBH and BCK are congruent by ASA postulate.

Congruent triangles have congruent corresponding sides, hence

BH ≅ CK.

5 0
3 years ago
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