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3 years ago

10

I am working on a test review. How can I write -1 <= 2x+7 <21 as two inequality statements? How do I join them with and/or

? This is for Algebra Compound Inequalities.

1 answer:

Tanya [424]3 years ago

6 0

Answer:

$( - 1 > 2x + 7)(2x + 7 < 21)$

Not sure about the and/or statement

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235+123= solve using a number line

Agata [3.3K]

The answer is 358....

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3 years ago

What expression is not equivalent to 2/3 *4.

Gelneren [198K]

Answer:

Step-by-step explanation:

Any expression that says (2 * 4)/3 is correct, even though there is a shift in the brackets. That means that A, B, C are all the same thing, and all are correct.

Only D is lying.

2*(1/3) = 2/3

4*(1/3) = 4/3 Now we have to add them

2/3 + 4/3 = 6/3 = 2

To show what the real answer is, try A

2*4/3

8/3 is the actual answer.

4 0

2 years ago

A village was founded four hundred years ago by a group of 20 people. In this village, the population triples every one hundred

Paha777 [63]

Answer:

Step-by-step explanation:

Treat this like compound interest: Use A = P(1 + r)^t.

Here, P is the initial population and A is 3 times that, or 3P. Since P = 20 people, 3P = 60 people,

and this population is reached after 100 years.

We need to determine r, substitute its value into the formula A = P(1 + r)^t, and then determine the population of the village after 400 years.

60 = 20(1 + r)^100

Simplifying, 3 = (1 + r)^100.

Taking the natural log of both sides,

ln 3 = 100 ln (1 + r), or

ln 3

ln (1 + r) = ---------------

100

= 1.0986 / 100 = 0.01986

We must solve this for r. Raising e to the power ln (1 + r), on the left side of an equation, and raising e to the power 0. 01986 on the right side, we get:

1 + r = 3, so r must = 2.

Now find the pop of the village today. Use the same equation: A = P (1+r)^t.

A = 20(1 +2)^4 (hundreds),

or

A = 20(3)^4, or

A = 81

The population after 400 years is 81.

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2 years ago

18 8/9 - 9 1/6 show your work

Genrish500 [490]

Answer:

8 17/18 or 8.945

Step-by-step explanation:

2/2x165/9-55/6x3/3

326/18-165/18

161/18= 8 17/18

5 0

3 years ago

1. What is the solution of n² – 49 = 0? (1 point)

maks197457 [2]

Answer:

1. $n=\pm7$

2. $x=\pm7i[tex]3.[tex]\pm 12x =a$

4. 22

Step-by-step explanation:

1. $n^2 - 49 = 0$

$n^2 =49$

$n= \sqrt{49}$

$n=\pm7$

2. $x^2+64 = 0$

$x^2 =-64$

$x= \sqrt{49}$

$x=\pm7i[tex]3. Area of square = [tex]a^{2}$

Where a is the side

We are given an area o square = $144x^{2}$

So, $144x^{2}=a^{2}$

$\sqrt{144x^{2}}=a$

$\pm 12x =a$

4. We are given that the solution of quadratic equation -9 and 9

So, equation becomes:

$(x+9)(x-9)=0$

$x^2- 81=0$

So, in the given equation $x^2 + z = 103$

z should be the number from which if we subtract 103 so we get 81

Substitute z = 22

$x^2 + 22 = 103$

$x^2 + 22 -103=0$

$x^2 -81=0$

Thus the value of z is 22

8 0

3 years ago

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