Answer:
Step-by-step explanation:
perp. 4/3
y - 4 = 4/3(x - 3)
y - 4 = 4/3x - 4
y = 4/3x
Answer:
B
Step-by-step explanation:
tan45° = 10/x
=> 1 = 10/x
=> x = 10
=> y^2 = 10^2 + 10^2 = 200
=> y = 100*sqrt(2)
=> b
Answer:
A one-sided, two-sample t-test
Step-by-step explanation:
In the context, the values of the population standard deviations are not provided, therefore we have to use the two sample t tests for the difference between the two population means. This claim for the hypothesis or the alternative hypothesis is given as the average selling price is more for the first group of the sample size of 21 as compared to the other group of the sample size of 87. Therefore, we perform an one sided test or the upper tailed test.
The total mass, in kilograms, of the nails the carpenter bought is 3.9 kilogram
<h3><u>Solution:</u></h3>
Given that carpenter bought 750 nails
Each nail has a mass of
kilogram
To find: total mass, in kilograms of the nails bought
The total mass of the nails bought can be found out by multiplying number of nails bought by mass of each nail
Number of nails bought = 750
Mass of each nail =
kilogram
total mass of the nails bought = number of nails bought x mass of each nail


Thus the total mass of nails bought is 3.9 kilogram
Answer:
- m∠A ≈ 53.13°
- m∠B ≈ 73.74°
- m∠C ≈ 53.13°
Step-by-step explanation:
An altitude to AC bisects it and creates two congruent right triangles. This lets you find ∠A = ∠C = arccos(6/10) ≈ 53.13°.
Since the sum of angles of a triangle is 180°, ∠B is the supplement of twice this angle, so is about 73.74°.
m∠A = m∠C ≈ 53.13°
m∠B ≈ 73.74°
_____
The mnemonic SOH CAH TOA reminds you of the relation between the adjacent side, hypotenuse, and trig function of an angle:
Cos = Adjacent/Hypotenuse
If the altitude from B bisects AC at X, triangle AXB is a right triangle with side AX adjacent to the angle A, and side AB as the hypotenuse. AX is half of AC, so has length 12/2 = 6, telling you the cosine of angle A is AX/AB = 6/10.
A diagram does not have to be sophisticated to be useful.