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3 years ago

Hellpppppp!!!!!!! me pleaseeeee

Mathematics

1 answer:

zloy xaker [14]3 years ago

7 0

Answer:

The correct answer is d because m<3 is supplementary (meaning the angles add to 180°) to m<4. It's easy to spot two supplementary angles because they will form a linear line.

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I need help I do not get it please help

zhenek [66]

The first fraction is 4 over 6, right? hopefully it is, i cant really see.
Basically, you can divide 4 and 6 by the same number, they have factors in common. They're both multiples of 2. if you divide them both by 2, you get 2 over 3, which is the <em>exact same amount as 4 over 6, it's just written in a simpler way. </em>the first answer is 2 over 3.

What numbers can both 10 and 15 be divided by? give it a try.

4 0

3 years ago

Read 2 more answers

How do I do this problem???

Andreas93 [3]

A. there are 72 right-handed batters. this is what you do:
5:8
45:?
45/5=9
8*9=72 so
45:72
b. there are 45 left-handed batters
the total batters is 45+72=117
so the ratio of left-handed to the total batter would be 45:117
hope this helps :)

4 0

2 years ago

Can you simpilfy the expression of: 6x-8y-5x+3y<br><br>Answer in comments

Romashka-Z-Leto [24]

Answer:

Hey there. Ur answer is x-5y

Step-by-step explanation:

5 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]

8 0

2 years ago

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Find the value of y and z.

My name is Ann [436]

Answer:

z= 125 because they are congruent and y = 55 because they are congruent

I hope this is good enough:

3 0

2 years ago

Hellpppppp!!!!!!! me pleaseeeee​

Hellpppppp!!!!!!! me pleaseeeee