Ask question

Ask question

All categories

3 years ago

7

WHAT IS THE ANSWER TO THIS QUESTION?

1 answer:

Triss [41]3 years ago

5 0

Answer:

this is a test I will not help you

Step-by-step explanation:

bye bye

You might be interested in

Find the area of the trapeziod 25 ft 20 ft 15 ft

Natalka [10]

The area of the trapeziod is 337.5

3 0

3 years ago

determine the maximum or minimum of the quadratic function. express your answer in the form (x,y) and using decimals rounded to

kolezko [41]

We are given the following quadratic equation

$f(x)=2x^2+7x-10$

The vertex is the maximum/minimum point of the quadratic equation.

The x-coordinate of the vertex is given by

$h=-\frac{b}{2a}$

Comparing the given equation with the general form of the quadratic equation, the coefficients are

a = 2

b = 7

c = -10

$h=-\frac{b}{2a}=-\frac{7}{2(2)}=-\frac{7}{4}=-1.75$

The y-coordinate of the vertex is given by

$\begin{gathered} f(x)=2x^2+7x-10 \\ f(-1.75)=2(-1.75)^2+7(-1.75)-10 \\ f(-1.75)=2(3.0625)^{}-12.25-10 \\ f(-1.75)=6.125^{}-12.25-10 \\ f\mleft(-1.75\mright)=-16.13 \end{gathered}$

This means that we have a minimum point.

Therefore, the minimum point of the given quadratic equation is

$(-1.75,-16.13)$

7 0

1 year ago

Read 2 more answers

Calculate the angle of depression from the

Verizon [17]

Answer: $52.6^{\circ}$

Step-by-step explanation:

Let the angle of depression be x.

$\sin x=\frac{27}{34}\\\\x=\sin^{-1} \left(\frac{27}{34} \right) \approx 52.6^{\circ}$

7 0

2 years ago

Verify by substitution whetherthe given functions are solutions of the given DE. Primes denote derivatives with respect to x.y!!

julia-pushkina [17]

Complete Question

The complete question is shown on the first uploaded

Answer:

$y_1$ is not a solution of the differential equation

$y_2$ is not a solution of the differential equation

$y_3$ is not a solution of the differential equation

Step-by-step explanation:

The differential equation given is $y'' + y' = cos2x$

Let consider the first equation to substitute

$y_1 = cosx +sinx$

$y_1' = -sinx +cosx$

$y_1'' = -cosx -sinx$

So

$y_1'' - y_1' = -cosx -sinx -sinx +cosx$

$y_1'' + y_1' = -2sinx$

So

$-2sinx \ne cos2x$

This means that $y_1$ is not a solution of the differential equation

Let consider the second equation to substitute

$y_2 = cos2x$

$y_2' = -2sin2x$

$y_2'' = -4cos2x$

So

$y_2'' + y_2' = -4cos2x-2sin2x$

So

$-4cos2x-2sin2x \ne cos2x$

This means that $y_2$ is not a solution of the differential equation

Let consider the third equation to substitute

$y_3 = sin 2x$

$y_3' = 2cos 2x$

$y_3'' = -4sin2x$

So

$y_3'' + y_3' = -4sin2x - 2cos2x$

So

$-4sin2x - 2cos2x \ne cos2x$

This means that $y_3$ is not a solution of the differential equation

6 0

3 years ago

What does the mode represent in a set of numbers?

BartSMP [9]

Answer:

The number which appears most often in a set of numbers.

4 0

3 years ago

Read 2 more answers