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2 years ago

Can someone help me on question 14. Please show steps

Mathematics

1 answer:

zavuch27 [327]2 years ago

5 0

Answer:

$0.5a+7.5b^2$

Step-by-step explanation:

$4b^2+3.5a-2a+3.5b^2-a$

Group similar terms together:

$3.5a-2a-a+4b^2+3.5b^2$

Add the term $b^2$ :

$3.5a-2a-a+7.5b^2$

Subtract the term a:

$0.5a+7.5b^2$

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f1(x) = ex, f2(x) = e−x, f3(x) = sinh(x) g(x) = c1f1(x) + c2f2(x) + c3f3(x) Solve for c1, c2, and c3 so that g(x) = 0 on the int

eimsori [14]

Answer:

(C1, C2, C3) = (K, K, -2K)

For K in the interval (-∞, ∞)

Step-by-step explanation:

Given

f1(x) = e^x

f2(x) = e^(-x)

f3(x) = sinh(x)

g(x) = 0

We want to solve for C1, C2 and C3, such that

C1f1(x) + C2f2(x) + C3f3(x) = g(x)

That is

C1e^x + C2e^(-x) + C3sinh(x) = 0

The hyperbolic sine of x, sinh(x), can be written in its exponential form as

sinh(x) = (1/2)(e^x + e^(-x))

So, we can rewrite

C1e^x + C2e^(-x) + C3sinh(x) = 0

C1e^x + C2e^(-x) + C3(1/2)(e^x + e^(-x)) = 0

So we have

(C1 + (1/2)C3)e^x + (C2 + (1/2)C3)e^(-x) = 0

We know that

e^x ≠ 0, and e^(-x) ≠ 0

So we must have

(C1 + (1/2)C3) = 0...........................(1)

and

(C2 + (1/2)C3) = 0..........................(2)

From (1)

2C1 + C3 = 0

=> C3 = -2C1.................................(3)

From (2)

2C2 + C3 = 0

=> C3 = -2C2................................(4)

Comparing (3) and (4)

2C1 = 2C2

=> C2 = C1

Let C1 = C2 = K

C3 = -2K

(C1, C2, C3) = (K, K, -2K)

For K in the interval (-∞, ∞)

3 0

2 years ago

Two triangles are similar with a scale factor of 3. The preimage triangle has a base of 7 inches and a height of 10 inches. What

vaieri [72.5K]

Answer:

<u>The area of the larger triangle is 315 inches²</u>

Step-by-step explanation:

1. Let's review all the information provided for answering the questions properly:

Pre-image triangle has a base of 7 inches and a height of 10 inches

Scale used : Factor of 3

2. What is the area of the larger triangle??

For calculating the area of the larger triangle, we use the scale this way:

Base = 7 inches * 3

21 inches

Height = 10 inches * 3

30 inches

Area of the larger triangle = 1/2 (Base * Height)

Area of the larger triangle = 1/2 (21 * 30)

<u>Area of the larger triangle = 630/2 = 315 inches²</u>

6 0

2 years ago

Suppose you roll a six-sided 50 times and calculate the mean roll, x .

Vitek1552 [10]

Answer:

0.07

a. the distribution will be a normal distribution.

c. we would suspicious ins there is a 2 % chance of getting the required value.

Step-by-step explanation:

Let the number of times, t be = 50

Assuming that the die is fair

standard deviation = 1.71

mean = 3.5

suppose we want tp find the probability of a 2 showing. The solution becomes:

probability = $\frac{2}{6} = \frac{1}{3}$

c. the mean of rolls will be 0.07

8 0

2 years ago

There are 100 freshman students at a school. The probability that a freshman plays a sport is 0.55. The probability that a fresh

Lina20 [59]

The probability that a freshman student at this school plays either a sport or a musical instrument is 0.74

Step-by-step explanation:

The addition rules in probability are:

P(A or B) = P(A) + P(B) ⇒ mutually exclusive (events cannot happen at the same time)
P(A or B) = P(A) + P(B) - P(A and B) ⇒ non-mutually exclusive (if they have at least one outcome in common)

∵ The probability that a freshman plays a sport is 0.55

∴ P(sport) = 0.55

∵ The probability that a freshman plays a musical instrument is 0.34

∴ P(music) = 0.34

∵ The probability that a freshman plays both a sport and a musical

instrument is 0.15

∴ P(sport and music) = 0.15

To find the probability that freshman student at this school plays either a sport or a musical instrument use the second rule above because it is non-mutually exclusive

∵ P(sport or music) = P(sport) + P(music) - P(sport and music)

∴ P(sport or music) = 0.55 + 0.34 - 0.15

∴ P(sport or music) = 0.74

The probability that a freshman student at this school plays either a sport or a musical instrument is 0.74

Learn more:

You can learn more about the probability in brainly.com/question/9178881

#LearnwithBrainly

5 0

2 years ago

Let $\overline{XY}$ be a tangent to a circle, and let $\overline{XBA}$ be a secant of the circle, as shown below. If $AX = 15$ a

KiRa [710]

Answer:

AB=5.4 units

Step-by-step explanation:

We using the theorem of intersecting tangent and secant to solve this.

By this theorem:

$XY^2=XB \times AX\\9^2=15(15-XB)\\81=225-15XB\\15XB=144\\XB=9.6\\$Therefore:\\AB=AX-XB\\AB=15-9.6\\AB=5.4$ units$

8 0

2 years ago