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2 years ago

14) ΔEFG ~ ΔLMN. Find the ratio of ΔEFG to ΔLMN. A) 1:4 B) 1:2 C) 2:1 D) 4:1

Mathematics

2 answers:

Inessa05 [86]2 years ago

6 0

Answer:

I think the answer is 2:1 but i can't be sure because there is no picturewith the corresponding numbers.

Paha777 [63]2 years ago

5 0

The solution is 2:1. To find the scale factor, simplify $\frac{FG}{MN}$ .

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Please answerrrrrr

Alona [7]

Answer:

x= -48 y =48

Step-by-step explanation:

-46 and 2 is -48 numbers away

3 0

2 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

C. $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

From the above expression, the power of (5) is 2

Express 2 as 6 - 4

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

Check the list of options for the expression on the left hand side

The correct answer is $(5-\frac{1}{2})^6$

3 0

2 years ago

Find the area of the shaded figure. Give answer in decimal to the hundredth place.

Firdavs [7]

Answer:

13.5

Step-by-step explanation:

72/(16/3)

6 0

2 years ago

Which one is it A. B. C. or D.

Anna35 [415]

the correct one is A :)

7 0

3 years ago

A production process produces 2% defective parts. a sample of 5 parts from the production is selected. what is the probability t

Lana71 [14]

Answer:

Probability of a sample that contains exactly two defective parts is .0037 or .37%

Step-by-step explanation:

As we know if P is the probability of achieving k results in n trials then probability formula is P = $\binom{n}{k}p^{K}q^{n-k}$

In this formula n = number of trials

k = number of success

(n-k) = number of failures

p = probability of success in one trial

q = (1-p) = probability of failure in one trial

In this sum n = 5

k = 2

number failures (n-k) = (5-2) = 3

p = 2% which can be written as .02

q = 98% Which can be written as .98

Now putting these values in the formula

P = $\binom{5}{2}(.02)^{2}(.98)^{5-2}$

P = $\binom{5}{2}(.02)^{2}(.98)^{3}$

$\binom{5}{2}= 5!/3!2!$

= 5×4×3×2×1/3×2×1×2×1

= 5×2 =10

P = 10×(.02)²×(.98)³

= .0037 or .37%

4 0

3 years ago

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