Please help now im taking a test

A mass weighting 64 pounds, attached to the end of a spring and stretches the spring 0.32 foot. There is no damping and no exter

scoundrel [369]

Answer:

Step-by-step explanation:

Check attachment for solution

Download pdf

8 0

2 years ago

Given the functions below, find f(x) - g(x) f(x) = 3x^2 + 2x + 1 g(x) = x^2 - 6x + 3

saveliy_v [14]

Answer:

Here is your answer.....

Hope it helps....

5 0

2 years ago

In ΔABC (m∠C = 90°), the points D and E are the points where the angle bisectors of ∠A and ∠B intersect respectively sides BC an

Airida [17]

This is a little long, but it gets you there.

ΔEBH ≅ ΔEBC . . . . HA theorem
EH ≅ EC . . . . . . . . . CPCTC
∠ECH ≅ ∠EHC . . . base angles of isosceles ΔEHC
ΔAHE ~ ΔDGB ~ ΔACB . . . . AA similarity
∠AEH ≅ ∠ABC . . . corresponding angles of similar triangle
∠AEH = ∠ECH + ∠EHC = 2∠ECH . . . external angle is equal to the sum of opposite internal angles (of ΔECH)
ΔDAC ≅ ΔDAG . . . HA theorem
DC ≅ DG . . . . . . . . . CPCTC
∠DCG ≅ ∠DGC . . . base angles of isosceles ΔDGC
∠BDG ≅ ∠BAC . . . .corresponding angles of similar triangles
∠BDG = ∠DCG + ∠DGC = 2∠DCG . . . external angle is equal to the sum of opposite internal angles (of ΔDCG)
∠BAC + ∠ACB + ∠ABC = 180° . . . . sum of angles of a triangle
(∠BAC)/2 + (∠ACB)/2 + (∠ABC)/2 = 90° . . . . division property of equality (divide equation of 12 by 2)
∠DCG + 45° + ∠ECH = 90° . . . . substitute (∠BAC)/2 = (∠BDG)/2 = ∠DCG (from 10 and 11); substitute (∠ABC)/2 = (∠AEH)/2 = ∠ECH (from 5 and 6)
This equation represents the sum of angles at point C: ∠DCG + ∠HCG + ∠ECH = 90°, ∴ ∠HCG = 45° . . . . subtraction property of equality, transitive property of equality. (Subtract ∠DCG+∠ECH from both equations (14 and 15).)

5 0

2 years ago

What is the -2 squared -2

oksano4ka [1.4K]

Answer: the answer is negative one over four

-1/4

6 0

3 years ago

The amount of time all students in a very large undergraduate statistics course take to complete an examination is distributed c

Anestetic [448]

Answer:

a) The mean is $\mu = 60$

b) The standard deviation is $\sigma = 9$

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

The probability a student selected at random takes at least 55.50 minutes to complete the examination equals 0.6915.

This means that when X = 55.5, Z has a pvalue of 1 - 0.6915 = 0.3085. This means that when $X = 55.5, Z = -0.5$