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3 years ago

Based on the graph, between which two values of x is a zero of g located?

Mathematics

1 answer:

mel-nik [20]3 years ago

5 0

Answer:

Between 4 and 5

Step-by-step explanation:

Given

See attachment for graph

Required

Between which points is g(x) = 0

From the attachment, we can see that the graph crosses the x-axis at 4... and -1...

This means that

g(x) = 0 between 4 and 5

g(x) = 0 between -1 and -2

Going by the available options,

g(x) = 0 between 4 and 5

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If your momma's mass is 405.45 lbs and the elevator holds up to 4,505 lbs how many mommas can you fit in the elevator?

aleksandr82 [10.1K]

Answer:

11 mommas!

Step-by-step explanation:

4505 divided by 405.45 is 11.11... Of course, you can't fit 11.11... mommas, so the answer is just 11.

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3 years ago

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a mixture of medicine consists of three vitamins A, B and C,if the ratio of the vitamins should be 5:1:3 and the total mixture i

Sergio039 [100]

Answer:

A≈22 | B≈4.4 | C≈13.2

Step-by-step explanation:

5 + 1 + 3= 9

40/9≈ 4.444…

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3 years ago

Help with this question (100 points)

Masteriza [31]

Answer:

y = $\frac{\pi}{3}$

Step-by-step explanation:

using the addition formula for sine

sin(x + y) = sinxcosy + cosxsiny

then

sinxcosy + cosxsiny = $\frac{1}{2}$ sinx + $\frac{\sqrt{3} }{2}$ cosx

for the 2 sides to be equal , then

cosy = $\frac{1}{2}$ and siny = $\frac{\sqrt{3} }{2}$

then

y = $cos^{-1}$ ( $\frac{1}{2}$ ) = $\frac{\pi }{3}$

and

y = $sin^{-1}$ ( $\frac{\sqrt{3} }{2}$ ) = $\frac{\pi }{3}$

thus y = $\frac{\pi }{3}$

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2 years ago

Solve (x + 3)2 + (x + 3) – 2 = 0. Let u = Rewrite the equation in terms of u. (u2 + 3) + u – 2 = 0 u2 + u – 2 = 0 (u2 + 9) + u –

Verdich [7]

Answer:

The solutions of the original equation are x=-5 and x=-2

Step-by-step explanation:

we have

$(x+3)^2+(x+3)-2=0$

Let

$u=(x+3)$

Rewrite the equation

$(u)^2+(u)-2=0$

Complete the square

$u^2+u=2$

$u^2+u+1/4=2+1/4$

$u^2+u+1/4=9/4$

rewrite as perfect squares

$(u+1/2)^2=9/4$

square root both sides

$(u+1/2)=\pm\frac{3}{2}$

$u=(-1/2)\pm\frac{3}{2}$

$u=(-1/2)+\frac{3}{2}=1$

$u=(-1/2)-\frac{3}{2}=-2$

the solutions are

u=-2,u=1

<em>Alternative Method</em>

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$(u)^2+(u)-2=0$

$a=1\\b=1\\c=-2$

substitute in the formula

$u=\frac{-1\pm\sqrt{1^{2}-4(1)(-2)}} {2(1)}$

$u=\frac{-1\pm\sqrt{9}} {2}$

$u=\frac{-1\pm3} {2}$

$u=\frac{-1+3} {2}=1$

$u=\frac{-1-3} {2}=-2$

the solutions are

u=-2,u=1

<em>Find the solutions of the original equation</em>

For u=-2

$-2=(x+3)$ ----> $x=-2-3=-5$

For u=1

$1=(x+3)$ ----> $x=1-3=-2$

therefore

The solutions of the original equation are

x=-5 and x=-2

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Answer: that not a real question it cant be answered has to be missing something

Step-by-step explanation:)

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