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3 years ago

In ∆ABC if AB = 6 cm , BC = 8cm, AC = 10 cm then value of ∠B is ________

Mathematics

1 answer:

Fudgin [204]3 years ago

5 0

Answer:

90 degrees

Step-by-step explanation:

B is the corner and angle opposite of the side AC.

so, AC is becoming side c, and the other two are a and b (it does not matter which is which).

we use the enhanced Pythagoras formula for general triangles

c² = a² + b² - 2ab×cos(C)

in our example the angle C is named B.

but other than that we simply calculate

10² = 6² + 8² - 2×6×8×cos(B)

100 = 36 + 64 - 96×cos(B)

100 = 100 - 96×cos(B)

0 = -96×cos(B)

cos(B) = 0

B = 90 degrees

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3 years ago

A fitness club charges its members a one time fee of 40$ and a monthly rate of 25$ what is this in q expression

joja [24]

Y = 25x + 40 is the expression.

6 0

3 years ago

Solve for q<br> (Please show all work thank you:))

Masja [62]

Answer:

q = ± $\frac{5}{4\sqrt{p} }$

Step-by-step explanation:

We are to solve for q in:

16pq² = 25

q² = $\frac{25}{16p}$

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2 years ago

A rectangular parking lot has a perimeter of 820 ft. The area of the parking lot measure SL 42,000 ft

horsena [70]

Answer:

a= 200

b = 210

Step-by-step explanation:

My assumption is, we have to find the length of sides of rectangle

Given

perimeter = 2a + 2b = 820 ft (i) (here a is smaller side and b is larger side)

area = a*b = 42,000 ft^2 (ii)

from eq (1)

2a + 2b = 820

=> 2(a+b) = 820

=> a+b = 820/2

=> a + b = 410

=> a = 410-b (iii)

putting the value of a in eq(ii), we get

(410-b) *b = 42,000

410b - b^2 = 42,000

0 = b^2 - 410b + 42000

b^2 - 410b + 42000 = 0

b^2- 200b- 210b + 42000 = 0

b(b-200)-210(b-200) = 0

(b-200)(b-210) = 0

b= 210 and b = 200

if b is larger side than b =210

By putting value of b in eq(iii),

a = 410 -210 = 200

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3 years ago

An educational psychologist wishes to know the mean number of words a third grader can read per minute. She wants to make an est

nirvana33 [79]

Answer:

The 80% confidence interval for the mean number of words a third grader can read per minute is between 40.4 wpm and 40.8 wpm.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.8}{2} = 0.1$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.1 = 0.99$ , so $z = 1.28$

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.28*\frac{4.1}{\sqrt{840}} = 0.2$

The lower end of the interval is the sample mean subtracted by M. So it is 40.6 - 0.2 = 40.4 words per minute.

The upper end of the interval is the sample mean added to M. So it is 40.6 + 0.2 = 40.8 words per minute.

The 80% confidence interval for the mean number of words a third grader can read per minute is between 40.4 wpm and 40.8 wpm.

3 0

3 years ago

In ∆ABC if AB = 6 cm , BC = 8cm, AC = 10 cm then value of ∠B is ________​

In ∆ABC if AB = 6 cm , BC = 8cm, AC = 10 cm then value of ∠B is ________