I don’t get this. The answer I chose is wrong.

A recycling station brought in 1/2 T of newspapers on Monday. On Tuesday, the station collected only half of what it did on Mond

Oxana [17]

Its b took the test Rita b duemwowms

8 0

2 years ago

Kal is decorating the outside edge of a square

Grace [21]

Answer: Option A
Take not that this is a square bulletin board thus area = Length Times Base. Notice how it says its a square as well? thus telling us that each side is of equal length

Squaring the value N would give us the value of the one side of the board

5 0

3 years ago

What is the diameter of a circular swimming pool with a radius of 8 feet?

Alenkasestr [34]

We know that,

diameter is twice the radius, I.e d = 2r

Now d = 2(8)ft = 16 feet

5 0

2 years ago

UDD Question 6 10 pts Solve for x. Round to the nearest tenth (one decimal place). Just type the number, no unit of measurement.

NISA [10]

Answer:

The answer is;

54.4

Step-by-step explanation:

The given parameter are;

Triangle ΔFGH = Right triangle

The length of segment $\overline {FH}$ = 48

The measure of ∠HFG = 28°

The measurement required = The measure of segment $\overline {FG}$ = x

The measure of ∠FHG = 90° angle opposite hypotenuse side of a right triangle

Therefore, ∠FGH = 180° - ∠HFG - ∠FHG = 180° - 90° - 28° = 62°

∠FGH = 62°

By sine rule, we have;

$\overline {FH}$ /(sin ∠FGH) = $\overline {FG}$ /(sin(∠FHG)

By substituting the known values, we have;

48/(sin 62°) = x/(sin(90°)

sin(90°) = 1, therefore, we have;

x/1 = x = 48/(sin 62°) = 54.4 (by rounding the answer to the nearest tenth)

x = 54.4

8 0

2 years ago

Given the function f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1, use intermediate theorem to decide which of the following intervals contai

marta [7]

$f(x) = x^4 + 3x^3 - 2x^2 - 6x - 1$

Lets check with every option

(a) [-4,-3]

We plug in -4 for x and -3 for x

$f(-4) = (-4)^4 + 3(-4)^3 - 2(-4)^2 - 6(-4) - 1= 55$

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

(b) [-3,-2]

We plug in -3 for x and -2 for x

$f(-3) = (-3)^4 + 3(-3)^3 - 2(-3)^2 - 6(-3) - 1= -1$

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

f(-2) is negative and f(-3) is negative. there is no value at x=c on the interval [-3,-2] where f(c)=0.

(c) [-2,-1]

We plug in -2 for x and -1 for x

$f(-2) = (-2)^4 + 3(-2)^3 - 2(-2)^2 - 6(-2) - 1= -5$

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

f(-2) is negative and f(-1) is positive. there is some value at x=c on the interval [-2,-1] where f(c)=0. so there exists atleast one zero on this interval.

(d) [-1,0]

We plug in -1 for x and 0 for x

$f(-1) = (-1)^4 + 3(-1)^3 - 2(-1)^2 - 6(-1) - 1= 1$

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

f(-1) is positive and f(0) is negative. there is some value at x=c on the interval [-1,0] where f(c)=0. so there exists atleast one zero on this interval.

(e) [0,1]

We plug in 0 for x and 1 for x

$f(0) = (0)^4 + 3(0)^3 - 2(0)^2 - 6(0) - 1= -1$

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

f(0) is negative and f(1) is negative. there is no value at x=c on the interval [0,1] where f(c)=0.

(f) [1,2]

We plug in 1 for x and 2 for x

$f(1) = (1)^4 + 3(1)^3 - 2(1)^2 - 6(1) - 1= -5$

$f(2) = (2)^4 + 3(2)^3 - 2(2)^2 - 6(2) - 1= 19$

f(-4) is positive and f(-3) is negative. there is some value at x=c on the interval [-4,-3] where f(c)=0. so there exists atleast one zero on this interval.

so answers are (a) [-4,-3], (c) [-2,-1], (d) [-1,0], (f) [1,2]

8 0

3 years ago

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