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2 years ago

12. Determine which ratio forms a proportion with 15/8 by using cross products.

Mathematics

2 answers:

Xelga [282]2 years ago

6 0

Answer to 12: C.

Explanation: By dividing 60 by 15 you get 4 and by dividing 32 by 8 you get 4. 15 and 8 both go into it evenly so that is the answer.

Answer to 14: B

Explanation: when you divide 236.25 by 7 you get 33.75 and then multiply that by 2 and you get 67.7 so you know 2 fluid ounces of milk contain 67.7 mg of calcium

sorry i couldnt answer 13

m_a_m_a [10]2 years ago

6 0

Answer:

Ratios that can form a proportion with 15/8 are those that are equivalent fractions. In this case, 60/32 can form a proportion with 15/8, because they are equivalent, if we simplify 60/32 we would have 15/8, because 60/4=15 and 32/4=8.

Therefore, the right answer is C.

Here we apply the same definition, a proportion is an equality between two fractions. In this case, the ratios that can form a proportion are A. 9/45, 8/40, because both fractions are equivalent to 0.2.

To solve this problem, we can apply the rule of three, if 7 ounces of milk provides 236.25 milligrams of calcium, how many ounces of milf would provide 67.5 milligrams of calcium.

$67.6 \ mgCa\frac{7 \ oz}{236.25 \ mgCa}= 2.00 oz$

Therefore, the right answer is B.

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Find the slope between the two points given. Then, use the slope and the first point to write the equation of the line in Point-

Vera_Pavlovna [14]

M(slope)= -5/4 point-slope(y-y1=m(x-x1)) so it would be y-5= -5/4(x+4)

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2 years ago

Read 2 more answers

Which three lengths could be the lengths

Fed [463]

Given:

The three lengths in the options.

To find:

Which three lengths could be the lengths of the sides of a triangle?

Solution:

If the sum of two smaller sides is greater than the largest side, then the three sides can form a triangle otherwise the triangle is not possible.

In option A, the three sides are 4 cm, 8 cm, 13 cm. Here, the largest side is 13 cm.

Sum of two smaller sides is:

$4+8=12$

$4+8$

So, the triangle is not possible.

In option B, the three sides are 13 cm, 11 cm, 23 cm. Here, the largest side is 23 cm.

Sum of two smaller sides is:

$13+11=24$

$13+11>23$

So, the triangle is possible.

In option C, the three sides are 5 cm, 18 cm, 12 cm. Here, the largest side is 18 cm.

Sum of two smaller sides is:

$5+12=17$

$5+12$

So, the triangle is not possible.

In option D, the three sides are 25 cm, 8 cm, 16 cm. Here, the largest side is 25 cm.

Sum of two smaller sides is:

$8+16=24$

$8+16$

So, the triangle is not possible.

Therefore, the correct option is B. The three lengths 13 cm, 11 cm, 23 cm could be the lengths of the sides of a triangle.

6 0

2 years ago

“encontrar la integral indefinida y verificar el resultado mediante derivación”

Oliga [24]

$I=\displaystyle\int\frac x{(1-x^2)^3}\,\mathrm dx$

Haz la sustitución:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{y^3}=\frac1{4y^2}+C=\frac1{4(1-x^2)^2}+C$

Para confirmar el resultado:

$\dfrac{\mathrm dI}{\mathrm dx}=\dfrac14\left(-\dfrac{2(-2x)}{(1-x^2)^3}\right)=\dfrac x{(1-x^2)^3}$

$I=\displaystyle\int\frac{x^2}{(1+x^3)^2}\,\mathrm dx$

Sustituye:

$y=1+x^3\implies\mathrm dy=3x^2\,\mathrm dx$

$\implies I=\displaystyle\frac13\int\frac{\mathrm dy}{y^2}=-\frac1{3y}+C=-\frac1{3(1+x^3)}+C$

(Te dejaré confirmar por ti mismo.)

$I=\displaystyle\int\frac x{\sqrt{1-x^2}}\,\mathrm dx$

Sustituye:

$y=1-x^2\implies\mathrm dy=-2x\,\mathrm dx$

$\implies I=\displaystyle-\frac12\int\frac{\mathrm dy}{\sqrt y}=-\frac12(2\sqrt y)+C=-\sqrt{1-x^2}+C$

$I=\displaystyle\int\left(1+\frac1t\right)^3\frac{\mathrm dt}{t^2}$

Sustituye:

$u=1+\dfrac1t\implies\mathrm du=-\dfrac{\mathrm dt}{t^2}$

$\implies I=-\displaystyle\int u^3\,\mathrm du=-\frac{u^4}4+C=-\frac{\left(1+\frac1t\right)^4}4+C$

Podemos hacer que esto se vea un poco mejor:

$\left(1+\dfrac1t\right)^4=\left(\dfrac{t+1}t\right)^4=\dfrac{(t+1)^4}{t^4}$

$\implies I=-\dfrac{(t+1)^4}{4t^4}+C$

4 0

3 years ago

The area of a circle with radius r is given by the formula A=πr2. If a circle has area 81π square inches, what is the radius (in

Ksenya-84 [330]

The Solution:

GIven:

$A=\pi r^2$

Required:

$Find\text{ }r\text{ when }A=81\pi in^2$

Substitute:

$\begin{gathered} \pi r^2=81\pi \\ \\ r^2=\frac{81\pi}{\pi}=81 \end{gathered}$

Take the square root of both sides:

$\begin{gathered} r=\sqrt{81}=\pm9 \\ \\ r=9in\text{ \lparen discard r=-9 since }r>0) \end{gathered}$

Answer:

9 inches

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1 year ago

goblinko [34]

4x6+24=48so that uis yo answer yoStep-by-step explanation:

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2 years ago