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2 years ago

15

Help mee pleaaseee!!!!!

2 answers:

iragen [17]2 years ago

5 0

C = is the correct answer

borishaifa [10]2 years ago

3 0

Answer: C) =

Step-by-step explanation: You can convert -14/4 from a fraction into a decimal. Take the denominator, 4, how many times does it go into the numerator, 14? 4 goes into 14 exactly 3.5 times. So, that means if the fraction was -14/4, you need to take that negative sign and put it in front of your decimal. Once you do that, you see that both the first and second number turn out to be the same number, -3.5. Therefore, your answer is C, as the numbers are equal to one another.

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Julie and her brother go to Atlanta to ride the SkyView Ferris wheel. It measures 200 feet in diameter.

Law Incorporation [45]

Answer:

Question 1: A)we need to use the CIRCUMFERENCE.

B) Circumference of a Circle = 2 π R

c)The distance traveled in 1 rotation = 628 ft

2(a)AREA of a CIRCLE = π R²

(b)Area the wheel takes up is 31,400 sq ft

Step-by-step explanation:

Question 1:

Diameter = 200 ft

Now,as Diameter = 2 x Radius

So, R = D/2 =200/2 = 100 ft

A) Distance around the wheel = The length of the boundary of the wheel.

So, we need to use the CIRCUMFERENCE.

B) Circumference of a Circle = 2 π R

Here R = Radius and use π = 3.14

C) The distance traveled in 1 rotation = 1 completer length of boundary

Now, Circumference = 2 π R = 2 x (3.14) x (100 ft) = 628 ft

Question 2:

A) Space the Ferris Wheel takes up = Area of the wheel

AREA of a CIRCLE = π R²

B) AREA of a CIRCLE = π R²

= (3.14) (100 ft ) (100 ft) = 31,400 sq ft

So, the area the wheel takes us is 31,400 sq ft

3 0

3 years ago

A box with a rectangular base and open top must have a volume of 128 f t 3 . The length of the base is twice the width of base.

noname [10]

Answer:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

Step-by-step explanation:

Given

$Volume = 128ft^3$

$L = 2W$

$Base\ Cost = \$9/ft^2$

$Sides\ Cost = \$6/ft^2$

Required

The dimension that minimizes the cost

The volume is:

$Volume = LWH$

This gives:

$128 = LWH$

Substitute $L = 2W$

$128 = 2W * WH$

$128 = 2W^2H$

Make H the subject

$H = \frac{128}{2W^2}$

$H = \frac{64}{W^2}$

The surface area is:

Area = Area of Bottom + Area of Sides

So, we have:

$A = LW + 2(WH + LH)$

The cost is:

$Cost = 9 * LW + 6 * 2(WH + LH)$

$Cost = 9 * LW + 12(WH + LH)$

$Cost = 9 * LW + 12H(W + L)$

Substitute: $H = \frac{64}{W^2}$ and $L = 2W$

$Cost =9*2W*W + 12 * \frac{64}{W^2}(W + 2W)$

$Cost =18W^2 + \frac{768}{W^2}*3W$

$Cost =18W^2 + \frac{2304}{W}$

To minimize the cost, we differentiate

$C' =2*18W + -1 * 2304W^{-2}$

Then set to 0

$2*18W + -1 * 2304W^{-2} =0$

$36W - 2304W^{-2} =0$

Rewrite as:

$36W = 2304W^{-2}$

Divide both sides by W

$36 = 2304W^{-3}$

Rewrite as:

$36 = \frac{2304}{W^3}$

Solve for $W^3$

$W^3 = \frac{2304}{36}$

$W^3 = 64$

Take cube roots

$W = 4$

Recall that:

$L = 2W$

$L = 2 * 4$

$L = 8$

$H = \frac{64}{W^2}$

$H = \frac{64}{4^2}$

$H = \frac{64}{16}$

$H = 4$

Hence, the dimension that minimizes the cost is:

$Width = 4ft$

$Height = 4ft$

$Length = 8ft$

8 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

Use the image to determine the value of “X”

PtichkaEL [24]

C. X= 18

ex:
4x-7=7x-61
(They equal eachother)
you solve and get 18

4 0

2 years ago

Please help me out!!!!!

Neko [114]

Given:

The zeros of the polynomial are $-2,\pm 5,4$ .

Degree = 4

Leading coefficient = 1

To find:

The polynomial.

Solution:

If c is a zero of a polynomial, then (x-c) must be a factor of the polynomial.

Here, -2,4,-5, 5, are zeros of the required polynomial, so (x+2), (x-4), (x+5), (x-5) are factors of required polynomial.

$p(x)=(x+2)(x-4)(x+5)(x-5)$

$p(x)=(x^2-4x+2x-8)(x^2-5^2)$ $[\because a^2-b^2=(a-b)(a+b)]$

$p(x)=(x^2-2x-8)(x^2-25)$

Using distributive property, we get

$p(x)=x^2(x^2-25)-2x(x^2-25)-8(x^2-25)$

$p(x)=x^4-25x^2-2x^3+50x-8x^2+200$

On combining like terms, we get

$p(x)=x^4-2x^3+(-25x^2-8x^2)+50x+200$

$p(x)=x^4-2x^3-33x^2+50x+200$

Here, the leading coefficient is 1. So, it is the required polynomial.

Therefore, the correct option is E.

3 0

2 years ago