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3 years ago

Find the area of a rectangle with the given base and height 5ft,6 in

Mathematics

1 answer:

anygoal [31]3 years ago

8 0

Answer:

360 inch

Step-by-step explanation:

Step 1: Converting

6inch = 6inch

5ft = 60inch

Step 2: Finding the area

Area of a rectangle = L x B

A = 6 x 60

A = 360inch

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kondor19780726 [428]

Answer:

23 degrees

Step-by-step explanation:

Find the diagram attached

The sum of angle on the straight line is 180°

103+3x+15+x+10 = 180

128+4x =180

4x = 180-128

4x = 52

x = 52/4

x = 13

The measure of the angle of the intersection between Derby Drive and Rosemont is x+10

= 13+10

= 23 degrees

Hence the measure of the angle is 23 degrees

5 0

3 years ago

1. A rectangle is 3/4 as wide as it is long. How

s344n2d4d5 [400]

Answer:

B. 6 inches

Step-by-step explanation:

To find how wide it is, multiply the length by 3/4

8(3/4)

= 6

The width of the rectangle in 6 inches.

So, the correct answer is B. 6 inches

5 0

2 years ago

What is the product of 84,90?

MAVERICK [17]

84•90 is 7,560.

ndheisn

3 0

3 years ago

What is this number in standard form? (5×1,000)+(9×1)+(2×1/10)+(6×1/1,000)

mr Goodwill [35]

Answer:

5009.206

Step-by-step explanation:

The answer is 5009.206. We see that the first portion of this question asks, "(5×1,000)+(9×1)". 5x1000 is 5000. 9x1 is 9. 5000+9= 5009. The second portion of this question asks for the numbers that go after the decimal. 2x1/10 is 2/10. 6x1/1000 is 6/1000. The decimal place stands for one. So we do .206. If the decimal stands for ones that means in this case the 2 is in the tenths place (like it should be since the question asks for 2/10) and the 6 is in the 1000 place (like it should be since the question asks for 6/1000). We add this up together and our answer is 5009.206. Hope this explanation helps! Happy learning!

5 0

3 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

Read 2 more answers