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1 year ago

Twice a number, increased by three, is between negative one and nine. Find all the numbers.

Mathematics

1 answer:

pashok25 [27]1 year ago

8 0

The solution set for the obtained inequality is {.......-5, -4, -3, -2, -1, 0, 1, 2}.

Given that, twice a number, increased by three, is between negative one and nine.

<h3>What is inequality?</h3>

The relation between two expressions that are not equal, employing a sign such as ≠ ‘not equal to’, > ‘greater than’, or < ‘less than’.

Let the unknown number be x.

Now, twice a number, increased by three = 2x+3

The given expression is between negative one and nine

-1 < 2x+3 < 9

-1 < 2x+3

Subtract 3 to both the side of the inequality

-1-3<2x+3-3

⇒ -4 < 2x

Divide 2 to both the side of the inequality

-2 < x

Now, 2x+3 < 9

Subtract 3 to both the side of the inequality

2x+3-3<9-3

⇒ 2x<6

Divide 2 to both the side of the inequality

x < 3

So, -2<x<3

Solution set = {.......-5, -4, -3, -2, -1, 0, 1, 2}

Therefore, the solution set for the obtained inequality is {.......-5, -4, -3, -2, -1, 0, 1, 2}.

To learn more about a inequality visit:

brainly.com/question/20383699.

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I need you to answer with a, b, c, d

solong [7]

To find the zeros of a quadratic fiunction given the equation you can use the next quadratic formula after equal the function to 0:

$\begin{gathered} ax^2+bx+c=0 \\ \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \end{gathered}$

For the given function:

$f(x)=2x^2-10x-3$ $x=\frac{-(-10)\pm\sqrt[]{(-10)^2-4(2)(-3)}}{2(2)}$ $x=\frac{10\pm\sqrt[]{100+24}}{4}$ $\begin{gathered} x=\frac{10\pm\sqrt[]{124}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2\cdot2\cdot31}}{4} \\ \\ x=\frac{10\pm\sqrt[]{2^2\cdot31}}{4} \\ \\ x=\frac{10\pm2\sqrt[]{31}}{4} \\ \end{gathered}$ $\begin{gathered} x_1=\frac{10}{4}+\frac{2\sqrt[]{31}}{4} \\ \\ x_1=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \end{gathered}$ $\begin{gathered} x_2=\frac{10}{4}-\frac{2\sqrt[]{31}}{4} \\ \\ x_2=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}$

Then, the zeros of the given quadratic function are:

$\begin{gathered} x=\frac{5}{2}+\frac{\sqrt[]{31}}{2} \\ \\ x_{}=\frac{5}{2}-\frac{\sqrt[]{31}}{2} \end{gathered}$

Answer: Third option

8 0

1 year ago

BRAINLIEST IF CORRECT PLEASE DONT ANSWER IF YOU DONT KNOW OR REPORTED 50 POINTS I NEED MY GRADES TO BE BETTER

weqwewe [10]

Answer:

B.166 11/12

Step-by-step explanation:

To handle this problem better, we can separate the fraction and integer portions to look at the whole situation better.

First, let's only look at the integers. Anthony flew up his drone 125 feet (translating to +125), allowed it to drop 14 feet (translating to -14), and then flew it up for 55 feet (translating to +55). Hence, only considering the integers, the total change in the drone's position would be (125 - 14 + 55) feet, which equates to 166 feet.

You'd probably have already noticed this answer is quite close to B. 166 11/12 feet, but let's keep on going for now.

On the fraction part, he flew his drone up 1/2 feet, allowed it to drop 1/3 feet, and then flew it up for 3/4 feet, translating to (1/2 - 1/3 + 3/4) feet.

Hence, after making all fractions share the same denominator, that same equation would change to (6/12 - 4/12 + 9/12) feet, which equals to 11/12 feet.

Finally, the total magnitude of the drone's position would be (166 + 11/12) feet, which is equal to B.166 11/12 feet.

Hope this helped!

6 0

3 years ago

Read 2 more answers

In ΔEFG, the measure of ∠G=90°, the measure of ∠E=6°, and GE = 71 feet. Find the length of EF to the nearest tenth of a foot.

djyliett [7]

Answer:68.2

Step-by-step explanation:

cos6=71/EF

cos6= 0.960170286650366

0.960170286650366= 71/EF

0.960170286650366*71=EF

EF= 68.2

3 0

3 years ago

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This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive an

sleet_krkn [62]

Answer:

-6re−r [sin(6θ) - cos(6θ)]

Step-by-step explanation:

the Jacobian is ∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

x = e−r sin(6θ), y = er cos(6θ)

δx/δθ = -6rcos(6θ)e−r sin(6θ), δx/δr = -sin(6θ)e−r sin(6θ)

δy/δθ = -6rsin(6θ)er cos(6θ), δy/δr = cos(6θ)er cos(6θ)

∂(x, y) /∂(r, θ) = δx/δθ × δy/δr - δx/δr × δy/δθ

= -6rcos(6θ)e−r sin(6θ) × cos(6θ)er cos(6θ) - [-sin(6θ)e−r sin(6θ) × -6rsin(6θ)er cos(6θ)]

= -6rcos²(6θ)e−r (sin(6θ) - cos(6θ)) - 6rsin²(6θ)e−r (sin(6θ) - cos(6θ))

= -6re−r (sin(6θ) - cos(6θ)) [cos²(6θ) + sin²(6θ)]

= -6re−r [sin(6θ) - cos(6θ)] since [cos²(6θ) + sin²(6θ)] = 1

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3 years ago

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saw5 [17]

1. x= 3/Y as a fraction btw

2. Same thing

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3 years ago

Twice a​ number, increased by ​three, is between negative one and nine. Find all the numbers.

Twice a number, increased by three, is between negative one and nine. Find all the numbers.