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3 years ago

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!! Simplify.

Mathematics

1 answer:

Finger [1]3 years ago

5 0

3x^3 - 7x^2 + 12 - 3x^3 -6x^2 - 10x

= -13x^2 - 10x + 12

Answer is B

-13x^2 - 10x + 12

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The length of a diagonal of a quadrilateral shaped field Is 48 m And the length of perpendicular dropped on it from the remainin

olchik [2.2K]

Answer:

$1008\text{ m}^2$

Step by step explanation:

Please find the attachment.

We have been given that the length of a diagonal of a quadrilateral shaped field Is 48 m. The length of perpendicular dropped on it from the remaining apposite vertices are 16 m and 26 m.

We can see from our attachment that diagonal AC divides our quadrilateral field into two triangular fields and AC will be the base of both triangles.

To find the area of our given quadrilateral we will find the areas of both triangles and add the areas.

Let us find area of triangle ABC with base 48 m and height 26 m.

$\text{Area of triangle}=\frac{1}{2}(\text{ Base}\times\text{ Height})$

$\text{Area of triangle ABC}=\frac{1}{2}\times 48\times 26$

$\text{Area of triangle ABC}=24\times 26$

$\text{Area of triangle ABC}=624$

Therefore, the area of triangle ABC will be 624 square meters.

Now let us find area of triangle ADC with base 48 m and height 16 m.

$\text{Area of triangle ADC}=\frac{1}{2}\times 48\times 16$

$\text{Area of triangle ADC}=24\times 16$

$\text{Area of triangle ADC}=384$

Therefore, the area of triangle ADC will be 384 square meters.

Now let us add areas of triangles ABC and ADC to get the area of our quadrilateral ABCD.

$\text{Area of quadrilateral ABCD}=384+624$

$\text{Area of quadrilateral ABCD}=1008$

Therefore, the area of quadrilateral shaped field will be 1008 square meters.

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4 years ago

Which is the better buy, Item A, selling at 5 and 2/3 pounds for $9.52 or Item B selling at 1 1/5 pounds for $1.98? Pls help !!

DanielleElmas [232]

Item B would be a better buy

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3 years ago

The temperature is -4 it increases to 8 how many degrees did the temperature increase

Oliga [24]

Answer:

12 degrees.

Step-by-step explanation:

8 - (-4) = 12. So, the temperature increased by 12 degrees.

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3 years ago

Consider an object moving along a line with the given velocity v. Assume t is time measured in seconds and velocities have units

Lady_Fox [76]

Answer:

a. the motion is positive in the time intervals: $[0,2)U(6,\infty)$

The motion is negative in the time interval: (2,6)

b. S=7 m

c. distance=71m

Step-by-step explanation:

a. In order to solve part a. of this problem, we must start by determining when the velocity will be positive and when it will be negative. We can do so by setting the velocity equation equal to zero and then testing it for the possible intervals:

$3t^{2}-24t+36=0$

so let's solve this for t:

$3(t^{2}-8t+12)=0$

$t^{2}+8t+12=0$

and now we factor it again:

(t-6)(t-2)=0

so we get the following answers:

t=6 and t=2

so now we can build our possible intervals:

$[0,2) (2,6) (6,\infty)$

and now we test each of the intervals on the given velocity equation, we do this by finding test values we can use to see how the velocity behaves in the whole interval:

[0,2) test value t=1

so:

$v(1)=3(1)^{2}-24(1)+36$

v(1)=15 m/s

we got a positive value so the object moves in the positive direction.

(2,6) test value t=3

so:

$v(1)=3(3)^{2}-24(3)+36$

v(3)=-9 m/s

we got a negative value so the object moves in the negative direction.

$(6,\infty)$ test value t=7

so:

$v(1)=3(7)^{2}-24(7)+36$

v(1)=15 m/s

we got a positive value so the object moves in the positive direction.

the motion is positive in the time intervals: $[0,2)U(6,\infty)$

The motion is negative in the time interval: (2,6)

b) in order to solve part b, we need to take the integral of the velocity function in the given interval, so we get:

$s(t)=\int\limits^7_0 {(3t^{2}-24t+36)} \, dt$

so we get:

$s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{7}_{0}$

which simplifies to:

$s(t)=[t^{3}-12t^{2}+36t]^{7}_{0}$

so now we evaluate the integral:

$s=7^{3}-12(7)^{2}+36(7)-(0^{3}-12(0)^{2}+36(0))$

s=7 m

for part c, we need to evaluate the integral for each of the given intervals and add their magnitudes:

[0,2)

s(t)=\int\limits^2_0 {(3t^{2}-24t+36)} \, dt

so we get:

$s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{2}_{0}$

which simplifies to:

$s(t)=[t^{3}-12t^{2}+36t]^{2}_{0}$

so now we evaluate the integral:

$s=2^{3}-12(2)^{2}+36(2)-(0^{3}-12(0)^{2}+36(0))$

s=32 m

(2,6)

s(t)=\int\limits^6_2 {(3t^{2}-24t+36)} \, dt

so we get:

$s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{6}_{2}$

which simplifies to:

$s(t)=[t^{3}-12t^{2}+36t]^{6}_{2}$

so now we evaluate the integral:

$s=6^{3}-12(6)^{2}+36(6)-(2^{3}-12(2)^{2}+36(2))$

s=-32 m

(6,7)

s(t)=\int\limits^7_6 {(3t^{2}-24t+36)} \, dt

so we get:

$s(t)=[\frac{3t^{3}}{3}-\frac{24t^{2}}{2}+36]^{7}_{6}$

which simplifies to:

$s(t)=[t^{3}-12t^{2}+36t]^{7}_{6}$

so now we evaluate the integral:

$s=7^{3}-12(7)^{2}+36(7)-(6^{3}-12(6)^{2}+36(6))$

s=7 m

and we now add all the magnitudes:

Distance=32+32+7=71m

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3 years ago

How are real numbers used to slove problems

Marina86 [1]

Answer:Real numbers are used in measurements of continuously varying quantities such as size and time, in contrast to the natural numbers 1, 2, 3, …, arising from counting. ... The real numbers include the positive and negative integers and fractions (or rational numbers) and also the irrational numbers.

Step-by-step explanation:

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4 years ago