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dedylja [7]
4 years ago
13

4(5 + 2x) -5 = 3(3x + 7)

Mathematics
1 answer:
Pie4 years ago
5 0

Answer:

x =  -1

Step-by-step explanation:

The first step is get rid of the numbers outside this (). So we will get

20+8x=9x+21 (multiplication) And we can get rid of one side of x by subtract 8x on both sides, an then we will get 20= x+21

And we can move the numbers to one side by subtract 21 on each side, and we will get -1=x

And rewrite it:

x = -1

That's the answer.

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Solve the following absolute value inequality.<br> |(5x-4)-3|&lt;0.01
12345 [234]

Answer:

<h2>Step-by-step explanation:</h2><h2>|(5x-4)-3|<0.01</h2><h2>so -0.01<(5x-4-3)<0.01</h2><h2>add 7</h2><h2>7-0.01<5x<7+0.01</h2><h2>6.99<5x<7.01</h2><h2>divide by 5</h2><h2>1.398<x<1.402</h2>
7 0
3 years ago
Bakery has bought 250 pounds of muffin dough. They want to make waffles or muffins in half-dozen packs out of it. Half a dozen o
Alexus [3.1K]

Answer:

250 batches of muffins and 0 waffles.

Step-by-step explanation:

-1

If we denote the number of batches of muffins as "a" and the number of batches of waffles as "b," we are then supposed to maximize the profit function

P = 2a + 1.5b

subject to the following constraints: a>=0, b>=0, a + (3/4)b <= 250, and 3a + 6b <= 1200. The third constraint can be rewritten as 4a + 3b <= 1000.

Use the simplex method on these coefficients, and you should find the maximum profit to be $500 when a = 250 and b = 0. So, make 250 batches of muffins, no waffles.

You use up all the dough, have 450 minutes left, and have $500 profit, the maximum amount.

6 0
3 years ago
Write out the first four terms of the series to show how the series starts. Then find the sum of the series or show that it dive
Nostrana [21]

Answer:

The first four terms of the series are

(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})

\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n} = 14.25

Step-by-step explanation:

We know that

Sum of convergent series is also a convergent series.

We know that,

\sum_{k=0}^\infty a(r)^k

If the common ratio of a sequence |r| <1 then it is a convergent series.

The sum of the series is \sum_{k=0}^\infty a(r)^k=\frac{a}{1-r}

Given series,

\sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}

=(9+3)+(\frac97+\frac35)+(\frac9{7^2}+\frac3{5^2})+(\frac9{7^3}+\frac3{5^3})+.......

The first four terms of the series are

(9+3),(\frac97+\frac35),(\frac9{7^2}+\frac3{5^2}),(\frac9{7^3}+\frac3{5^3})

Let

S_n=\sum_{n=0}^\infty \frac{9}{7^n}    and     t_n=\sum_{n=0}^\infty \frac{3}{5^n}

Now for S_n,

S_n=9+\frac97+\frac{9}{7^2}+\frac9{7^3}+.......

    =\sum_{n=0}^\infty9(\frac 17)^n

It is a geometric series.

The common ratio of S_n is \frac17

The sum of the series

S_n=\sum_{n=0}^\infty \frac{9}{7^n}

    =\frac{9}{1-\frac17}

    =\frac{9}{\frac67}

    =\frac{9\times 7}{6}

    =10.5

Now for t_n

t_n= 3+\frac35+\frac{3}{5^2}+\frac3{5^3}+.......

    =\sum_{n=0}^\infty3(\frac 15)^n

It is a geometric series.

The common ratio of t_n is \frac15

The sum of the series

t_n=\sum_{n=0}^\infty \frac{3}{5^n}

    =\frac{3}{1-\frac15}

    =\frac{3}{\frac45}

    =\frac{3\times 5}{4}

    =3.75

The sum of the series is \sum_{n=0}^\infty \frac9{7^n}+\frac{3}{5^n}

                                        = S_n+t_n

                                       =10.5+3.75

                                       =14.25

4 0
4 years ago
Tamara looked at the expression 35 + z &lt; 59. She believes that z can equal more than one value.
MakcuM [25]

Answer:

z<24

Step-by-step explanation:

It can't be more than one value. All you do is subtract 35 from both sides to isolate z.

Hope this helped!

4 0
3 years ago
Read 2 more answers
Determine how many cards, in an ordinary deck of fifty-two, fit the description. 4's and 5's
Rudiy27
Since there are four suits, with each having a 4 and 5, we have 2+2+2+2=2*4=8
8 0
3 years ago
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