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Stolb23 [73]
2 years ago
6

Triangle ABC Is similar to triangle DEF solve for r

Mathematics
1 answer:
Ymorist [56]2 years ago
4 0

since they're similair then similair sides will be proportional

CB is similair to EF

\frac{6}{8}86 = \frac{3}{4}43

same thing if you try with the hypotenuse AB similair to DE

\frac{13.5}{18}1813.5 = \frac{3}{4}43

So, if you multiply  r (DF) by 3/4  you will get the similair side AC

rx 3/4=12

r= 12÷ 3/4

<h3>r=16.</h3>

<h2>i HOPE IT'S HELP </h2><h3 />
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If g(x) = 4x + 1 and g(x) = -11, find x.
podryga [215]

Answer:

x = - 3

Step-by-step explanation:

g(x) = 4x + 1...... (1)

g(x) = -11.... (2)

From equations (1) & (2)

-11 = 4x + 1

-11 - 1 = 4x

-12 = 4x

-12/4 = x

-3 = x

x = - 3

3 0
2 years ago
Given vectors u = (−1, 2, 3) and v = (3, 4, 2) in R 3 , consider the linear span: Span{u, v} := {αu + βv: α, β ∈ R}. Are the vec
julia-pushkina [17]

Answer:

(2,6,6) \not \in \text{Span}(u,v)

(-9,-2,5)\in \text{Span}(u,v)

Step-by-step explanation:

Let b=(b_1,b_2,b_3) \in \mathbb{R}^3. We have that b\in \text{Span}\{u,v\} if and only if we can find scalars \alpha,\beta \in \mathbb{R} such that \alpha u + \beta v = b. This can be translated to the following equations:

1. -\alpha + 3 \beta = b_1

2.2\alpha+4 \beta = b_2

3. 3 \alpha +2 \beta = b_3

Which is a system of 3 equations a 2 variables. We can take two of this equations, find the solutions for \alpha,\beta and check if the third equationd is fulfilled.

Case (2,6,6)

Using equations 1 and 2 we get

-\alpha + 3 \beta = 2

2\alpha+4 \beta = 6

whose unique solutions are \alpha =1 = \beta, but note that for this values, the third equation doesn't hold (3+2 = 5 \neq 6). So this vector is not in the generated space of u and v.

Case (-9,-2,5)

Using equations 1 and 2 we get

-\alpha + 3 \beta = -9

2\alpha+4 \beta = -2

whose unique solutions are \alpha=3, \beta=-2. Note that in this case, the third equation holds, since 3(3)+2(-2)=5. So this vector is in the generated space of u and v.

4 0
2 years ago
A random sample of 49 statistics examinations was taken. the average score, in the sample, was 84 with a variance of 12.25. the
tester [92]

Since in this case we are only using the variance of the sample and not the variance of the real population, therefore we use the t statistic. The formula for the confidence interval is:

<span>CI = X ± t * s / sqrt(n)                      ---> 1</span>

Where,

X = the sample mean = 84

t = the t score which is obtained in the standard distribution tables at 95% confidence level

s = sample variance = 12.25

n = number of samples = 49

From the table at 95% confidence interval and degrees of freedom of 48 (DOF = n -1), the value of t is around:

t = 1.68

 

Therefore substituting the given values to equation 1:

CI = 84 ± 1.68 * 12.25 / sqrt(49)

CI = 84 ± 2.94

CI = 81.06, 86.94

 

<span>Therefore at 95% confidence level, the scores is from 81 to 87.</span>

6 0
3 years ago
Replace A with a number to make A/24 ≥ 1/4 a true statement.
antiseptic1488 [7]

Answer:

B. 7

Step-by-step explanation: 6/24= 1/4 so 7/24>1/4

7 0
3 years ago
Consider a prism ABCA'B'C' whose base is an equilateral triangle of side a. The projection of A' onto ABC is the centroid of ABC
Serggg [28]

Answer:

lsu is there to help you with this question

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
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