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AleksAgata [21]
3 years ago
13

What's the common denominator between 210 and 770

Mathematics
1 answer:
Kipish [7]3 years ago
7 0

Suppose you want to combine the fractions 13/210 and 29/770.  What's the LCD?

Factor both 210 and 770:  3*7*10 and 7*11*10.  Form the LCD using the common factors 7 and 10 and the unique factors 3 and 11.  The LCD is thus 3(7)(10)(11), or 2310.

Then 13/210 and 29/770 become 143/2310 and 87/2310.  We can add these because their denominators are the same.

So the desired LCD is 2310.

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10.
yan [13]

Answer:

D. y = -5^x – 27

Step-by-step explanation:

Find the negative reciprocal of the slope of the original line and use the point-slope formula  y − y 1 = m ( x − x 1 )  to find the line perpendicular to  − x + 5 y = 14 .

y = − 5 x − 27                         PERPENDICULAAR

...................................................................................................................................................

Find the slope of the original line and use the point-slope formula  y − y 1 = m ( x − x 1 )  to find the line parallel to  − x + 5 y = 14 .

y = 1 /5 x − 1                             PARALLEL

7 0
3 years ago
Classify the polynomial by its degree and the number of terms. 8x²- 3y + 7
katen-ka-za [31]

Answer:

2nd degree

Step-by-step explanation:

8 0
3 years ago
Winnie needs to spend $120 on two kinds of fancy cakes:
Ainat [17]

Answer:

Winnie needs to buy 6 slices of Coffee cake.

Step-by-step explanation:

12 * 5= $60 (chocolate cake)

120 - 60= 60

since we know the coffee cake costs $10(per slice) we do:

60/ $10= 6 slices of coffee cake.

3 0
3 years ago
Solve the initial value problem: y'(x)=(4y(x)+25)^(1/2) ,y(1)=6. you can't really tell, but the '1/2' is the exponent
goblinko [34]

Answer:

y(x)=x^2+5x

Step-by-step explanation:

Given: y'=\sqrt{4y+25}

Initial value: y(1)=6

Let y'=\dfrac{dy}{dx}

\dfrac{dy}{dx}=\sqrt{4y+25}

Variable separable

\dfrac{dy}{\sqrt{4y+25}}=dx

Integrate both sides

\int \dfrac{dy}{\sqrt{4y+25}}=\int dx

\sqrt{4y+25}=2x+C

Initial condition, y(1)=6

\sqrt{4\cdot 6+25}=2\cdot 1+C

C=5

Put C into equation

Solution:

\sqrt{4y+25}=2x+5

or

4y+25=(2x+5)^2

y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4}

y(x)=x^2+5x

Hence, The solution is y(x)=\dfrac{1}{4}(2x+5)^2-\dfrac{25}{4} or y(x)=x^2+5x

4 0
3 years ago
Gdsgdsgdsgdsgdsgdsgdsgdsgdsgdsgdsgdsgdsgdsgdsgds
KATRIN_1 [288]
Yes and thank you for the points
4 0
2 years ago
Read 2 more answers
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