Divide 406 by −14. A) −29 B) −34 C) 34 D) 44

Answer:

We have to complete the proof.

PQRS is a parallelogram          1. Given

2. [tex]\overline{PQ}\:is\:parallel\:to\:\overline{RS}[/tex]    Definition of a parallelogram

m∠1 ≅ m∠3                  4. Alternate interior angles are equal or

m∠2 ≅ m∠4                       congruent.

[tex]\overline{PQ}\:\cong\:\overline{RS}[/tex]      Opposite sides of a parallelogram are congruent

ΔPQT ≅ ΔRST     5. Triangle are congruent by Angle-Side-

                                 Angle congruent rule

[tex]\overline{PT}\:\cong\:\overline{RT}[/tex]   6. Corresponding parts of Congruent triangles

[tex]\overline{QT}\:\cong\:\overline{ST}[/tex]

7. PR and QS bisect each other at T          Definition of a bisector

To prove that PR and QS bisect each other at T in a parallelogram PQRS, we can show that PR and QS intersect at T and divide each other into equal segments.

To prove that PR and QS bisect each other at T, we can use the fact that PQRS is a parallelogram.

We can start by showing that PR and QS intersect at a point, say T. Since PQRS is a parallelogram, its opposite sides are congruent. Therefore, PR is congruent to QS.

Next, we can show that PR and QS divide each other into equal segments. We can do this by proving that triangle PTR is congruent to triangle STQ using the side-angle-side (SAS) congruence criteria.

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Find the function f whose gradient is equal to the vector function f, then evaluate the line integral of f along the given hyperbolic arc from (1,1) to (3,1/3). This involves calculus and the concept of vector fields.

This problem involves calculus, specifically the concept of vector fields and line integrals. First, we are asked to find a function f(x, y) such that its gradient is equal to a given vector function f(x, y). To find this scalar function, we need to perform the integral of each component of the vector function f in terms of x and y.

Then we need to evaluate the line integral of the vector function f along a curve c, which is given as an arc of the hyperbola y = 1/x. The definite integral from point (1,1) to point (3,1/3) should be calculated. This integral represents the work done by the field f along the curve c.

Keep in mind that when you perform a line integral, you are adding up the contributions of a field (in this case, the vector field f) along a path (here, the hyperbola). Therefore, the path must be parameterized, and the limits of integration should be the parameter values that correspond to the points (1, 1) and (3, 1/3).

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Answer: a. Bonds

Step-by-step explanation:

Bond is a kind of money investment issued by private corporations or government banks.

Buying a bond is like issuing a loan to the issuer. The issuer will pay back the face value of the issued loan along with interest money. It is issued for a fixed period of time.

Answer:

a) bonds

Step-by-step explanation:

was right

The area of composite figure of both Quarter circle and Parallelogram is 224inches.

"When a Circle is divided into four equal parts, then each parts of a circle is called Quarter of a Circle".

A parallelogram is "two dimensional geometrical shape, whose sides are parallel to each other".

According to the question,

Area of circle =πr²

Formula for Area of quarter Circle = (π/4)r²

Formula for Area of parallelogram = base × height = (b×h)

In order to find the area of composite figure

Formula for Area of composite figure =  Area of quarter Circle + Area of parallelogram

=  (π/4)r² + (b×h)  

= (3.14/4)(14)² + (14×5)

= 223.86inches ≈ 224inches

Hence, the area of composite figure of both Quarter circle and Parallelogram is 224 inches.

To learn more about Quarter circle and Parallelogram here

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Answer:

A = $35,198.32

Step-by-step explanation:

Use the formula to calculate compound interest:

A = P(1 + i)ⁿ

"A" for total amount after the time period

"P" for principal, or starting money

"i" for the interest rate in a compounding period

To calculate "i":

i = r / c

"n" for the number of compounding periods

To calculate "n":

n = tc

So, we can combine the formulas into:

[tex]A = P(1+\frac{r}{c})^{tc}[/tex]

"c" is the compounding periods in a year. (quarterly = 4)

We know:

P = 8000

r = 10% / 100 = 0.1

t = 15

c = 4

Substitute the information in the formula.

[tex]A = P(1+\frac{r}{c})^{tc}[/tex]

[tex]A = 8000(1+\frac{0.1}{4})^{15*4}[/tex]      Solve "i" and "n"

[tex]A = 8000(1+0.025)^{60}[/tex]        Solve inside the brackets

[tex]A = 8000(1.025)^{60}[/tex]         Do the exponent before multiplying by 8000

A = 35198.318             Exact answer

A ≈ 35198.32              Round to two decimal places for money

Therefore she will have $35,198.32 after 15 years.

Anisha will have approximately $25,266.57 after 15 years of compound interest at a rate of 10% compounded quarterly on her initial investment of $8,000.

Anisha invested $8,000 in an account that earns 10% interest compounded quarterly. To calculate the amount of money she will have in 15 years, we can use the formula for compound interest:

Amount = Principal * (1 + (Rate / Frequency))^(Frequency * Time)

Amount = $8,000 * (1 + (0.10 / 4))^(4 * 15)

Amount = $8,000 * (1 + 0.025)^60

Amount = $8,000 * (1.025)^60

Using a calculator, we find that the amount of money Anisha will have after 15 years is approximately $25,266.57

Answer:

4,000cm^2

Step-by-step explanation:

Final answer:

To find 1/9 as many as 3/5, you multiply 3/5 by 1/9 to get 3/45, which simplifies to 1/15.

Explanation:

To solve the student's question, we need to find what is 1/9 as many as 3/5. This involves multiplying the fraction 3/5 by the fraction 1/9. Here is the step-by-step calculation:

The calculation will look like this: (3/5) × (1/9) = 3/45 = 1/15.

Thus, 1/9 as many as 3/5 is 1/15.

Answer:

(8, 1.5)

Step-by-step explanation:

x^2 only touches the x axis when x=0 and it increases without bound on either side of 0.

-x^2-5 never touches the x axis reaching a maximum at (0,-5) and decreasing without bound on either side of 0

g(x) is f(x) inverted and moved down five units...

Answer:  The fraction of the children who have a dog also have a cat is [tex]\dfrac{1}{3}.[/tex]

Step-by-step explanation:  Given that [tex]\frac{1}{2}[/tex] of the children within the city limits have a dog,  [tex]\frac{1}{3}[/tex]  have a cat, and  [tex]\frac{1}{6}[/tex]  have a dog and a cat.

We are to find the fraction of those children who have a dog also have a cat.

Let, S denote the sample space of the given experiment.

Then, n(S) = 1.

Let, event A and event B represents the fractions of children having a dog and a cat respectively.

Then, according to the given information, we have

[tex]n(A)=\dfrac{1}{2},~~~n(B)=\dfrac{1}{3},~~~n(A\cap B)=\dfrac{1}{6}.[/tex]

The fraction of children who have a dog also have a cat is given by the conditional probability of event B given A.

That is, he required probability is

[tex]P(B/A)=\dfrac{P(B\cap A)}{P(A)}=\dfrac{\frac{1}{6}}{\frac{1}{2}}=\dfrac{1}{3}.[/tex]

Thus, the fraction of the children who have a dog also have a cat is [tex]\dfrac{1}{3}.[/tex]

A radical equation is just like any other equation, but having some algebraic expression of x's given inside a square-root symbol. It could be as simple as y = [tex]\sqrt{x}[/tex] or it could be any complex algebraic expression inside the radical sign.  

The main condition for existence of any radical equation is that we can not have negative values inside square-root i.e. the algebraic expression given inside radical sign must have non-negative values (zero or positive real values).

Answer:

it is 19

Step-by-step explanation:

I just took this test and got 100 :)