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3 years ago

WILL MARK BRAINLIEST IF GOTTEN RIGHT!!!!

Mathematics

1 answer:

nikdorinn [45]3 years ago

6 0

Answer:

C both line and point symmetry

Step-by-step explanation:

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After the first four games of a professional football season, three players scored a total of 80 touchdowns.

aleksandr82 [10.1K]

Andre Roberts scored a total of 34 touchdowns

An equation is an expression that shows the relationship between two or more variables and numbers.

The total touchdowns is 80.

Duke Johnson scored 14 of the touchdowns.

Josh McCown scored 40% of the touchdowns. = 40% of 80 = 0.4 * 80 = 32

Let x represent the number of touchdowns scored by Andre Roberts, hence:

x + 14 + 32 = 80

x + 46 = 80

x = 34

Andre Roberts scored a total of 34 touchdowns

Find more about equation at: brainly.com/question/2972832

7 0

2 years ago

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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]

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3 years ago

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Estimate ∫20(4x2−2x)dx= using a left Reimann Summ and 4 equal subintervals.

dolphi86 [110]

nice day folks eeeeeeeeeeeeeeeee

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3 years ago

I need the answer fast pls

ludmilkaskok [199]

A. C. D. E. F.

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3 years ago

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Try to say m without using Ur lips ( its impossible) if u can do that u will get a brainliest

Helga [31]

Answer:

It's the middle of the night and I'm sleeping at my school- I don't think they would like to hear. "pftpdfd" "pfff" "mpfp"(Also known as me trying to make the letter m sound without my lips) all night.

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2 years ago