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uysha [10]
3 years ago
10

Iam a number less then 3000. When u divide me by 32 my remainder is 30. When u divide me by 58 my remainder is 44. What number a

m I?
Mathematics
1 answer:
forsale [732]3 years ago
4 0
32a+30=YOU,
58b+44=YOU, a,b and YOU are integers; YOU<3000.

YOU can be 798, 1726, or 2654.



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What is the value of $x$ if $-\frac23(x-5) = \frac32(x+1)$?
Ipatiy [6.2K]

Answer:

x = (-29)/5

Step-by-step explanation:

Solve for x:

(2 (x - 5))/3 = (3 (x + 1))/2

Multiply both sides by 6:

(6×2 (x - 5))/3 = (6×3 (x + 1))/2

6/3 = (3×2)/3 = 2:

2×2 (x - 5) = (6×3 (x + 1))/2

6/2 = (2×3)/2 = 3:

2×2 (x - 5) = 3×3 (x + 1)

2×2 = 4:

4 (x - 5) = 3×3 (x + 1)

3×3 = 9:

4 (x - 5) = 9 (x + 1)

Expand out terms of the left hand side:

4 x - 20 = 9 (x + 1)

Expand out terms of the right hand side:

4 x - 20 = 9 x + 9

Subtract 9 x from both sides:

(4 x - 9 x) - 20 = (9 x - 9 x) + 9

4 x - 9 x = -5 x:

-5 x - 20 = (9 x - 9 x) + 9

9 x - 9 x = 0:

-5 x - 20 = 9

Add 20 to both sides:

(20 - 20) - 5 x = 20 + 9

20 - 20 = 0:

-5 x = 9 + 20

9 + 20 = 29:

-5 x = 29

Divide both sides of -5 x = 29 by -5:

(-5 x)/(-5) = 29/(-5)

(-5)/(-5) = 1:

x = 29/(-5)

Multiply numerator and denominator of 29/(-5) by -1:

Answer: x = (-29)/5

4 0
3 years ago
Step 1 5x- y= 4
valkas [14]

Answer: C, In Step 2 the expression for y should be subsituted in the other equation.

5 0
2 years ago
Divide the following quantiles in the given ratio £100 1:3 £80 3:5 £250 2:3:5
goblinko [34]

Answer:

Step-by-step explanation:

a.

£100

1:3

1+3=4

1/4×£100

=£25

3/4×£100

=£75

b.

£80

3:5

3+5=8

3/8×£80

=£30

5/8×£80

=£50

C.

£250

2:3:5

2+3+5=10

2/10×£250

=£50

3/10×£250

=£75

5/10×£250

=£125

6 0
3 years ago
Solve 73 make sure to also define the limits in the parts a and b
Aleks04 [339]

73.

f(x)=\frac{3x^4+3x^3-36x^2}{x^4-25x^2+144}

a)

\lim_{x\to\infty}f(x)=\lim_{x\to\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\lim_{x\to-\infty}f(x)=\lim_{x\to-\infty}(\frac{3+\frac{3}{x}-\frac{36}{x^2}}{1-\frac{25}{x^2}+\frac{144}{x^4}})=3\cdot\frac{1}{2}=3

b)

Since we can't divide by zero, we need to find when:

x^4-2x^2+144=0

But before, we can factor the numerator and the denominator:

\begin{gathered} \frac{3x^2(x^2+x-12)}{x^4-25x^2+144}=\frac{3x^2((x+4)(x-3))}{(x-3)(x-3)(x+4)(x+4)} \\ so: \\ \frac{3x^2}{(x+3)(x-4)} \end{gathered}

Now, we can conclude that the vertical asymptotes are located at:

\begin{gathered} (x+3)(x-4)=0 \\ so: \\ x=-3 \\ x=4 \end{gathered}

so, for x = -3:

\lim_{x\to-3^-}f(x)=\lim_{x\to-3^-}-\frac{162}{x^4-25x^2+144}=-162(-\infty)=\infty\lim_{x\to-3^+}f(x)=\lim_{x\to-3^+}-\frac{162}{x^4-25x^2+144}=-162(\infty)=-\infty

For x = 4:

\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty\lim_{x\to4^-}f(x)=\lim_{n\to4^-}\frac{384}{x^4-25x^2+144}=384(-\infty)=-\infty

4 0
10 months ago
2+3!!<br> Help!! Due in an hour!!
Zanzabum

Answer:

5

Step-by-step explanation:

I think

6 0
3 years ago
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