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Brilliant_brown [7]

3 years ago

13

How do you multiply a mixed fraction by a regular fraction

1 answer:

Dmitriy789 [7]3 years ago

3 0

Well, you first have to make the mixed fraction into an improper fraction. You then multiply the two fractions and ad the whole number into the total.

That's how i was taught back in the 4th grade... i'm not sure if it's correct but i hope it is-

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Rick has 7 cats. Carrie has 14 cats. How many fewer cats does Rick have

guapka [62]

Answer:

He has 7 fewer cats because

Step-by-step explanation:

14-7=7

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3 years ago

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3 years ago

Condense the following logs into a single log:

mamaluj [8]

QUESTION 1

The given logarithm is

$8\log_g(x)+5\log_g(y)$

We apply the power rule of logarithms; $n\log_a(m)=\log_(m^n)$

$=\log_g(x^8)+\log_g(y^5)$

We now apply the product rule of logarithm;

$\log_a(m)+\log_a(n)=\log_a(mn)$

$=\log_g(x^8y^5)$

QUESTION 2

The given logarithm is

$8\log_5(x)+\frac{3}{4}\log_5(y)-5\log_5(z)$

We apply the power rule of logarithm to get;

$=\log_5(x^8)+\log_5(y^{\frac{3}{4}})-\log_5(z^5)$

We apply the product to obtain;

$=\log_5(x^8\times y^{\frac{3}{4}})-\log_5(z^5)$

We apply the quotient rule; $\log_a(m)-\log_a(n)=\log_a(\frac{m}{n} )$

$=\log_5(\frac{x^8\times y^{\frac{3}{4}}}{z^5})$

$=\log_5(\frac{x^8 \sqrt[4]{y^3} }{z^5})$

7 0

3 years ago

Bezzdna [24]

Jerome bought C. $4\frac{7}{8}$ pounds of candy.

Step-by-step explanation:

Step 1:

First, we must convert the mixed fractions into improper fractions. To do that we multiply the whole number with the denominator and add with it the numerator while the denominator remains the same.

To convert this fraction $1\frac{1}{2} = \frac{(1)(2)+1}{2} =\frac{3}{2} .$

To convert this fraction $2\frac{3}{4} = \frac{(2)(4)+3}{4} =\frac{11}{4} .$

So Jerome bought $\frac{3}{2}$ pounds of jelly beans, $\frac{5}{8}$ pounds of gumballs and $\frac{11}{4}$ pounds of chocolate.

Step 2:

We add all the values to determine the total candy bought.

$\frac{3}{2} +\frac{5}{8} + \frac{11}{4} = \frac{(4(3)+1(5)+2(11))}{8} .$

$\frac{(4(3)+1(5)+2(11))}{8} = \frac{39}{8} .$

$\frac{39}{8}$ can also be written as $4\frac{7}{8}$ .

So Jerome bought $4\frac{7}{8}$ pounds of candy which is option C.

6 0

4 years ago

Find the area of the trapezoid below to the nearest tenth.

kondor19780726 [428]

Answer:

Area of trapezoid = 67.6 square units

Step-by-step explanation:

Area of a trapezoid is given by the expression,

Area of the trapezoid = $\frac{1}{2}(b_1+b_2)h$

Here, $b_1$ and $b_2$ are the parallel sides of the given trapezoid.

And ' $h$ ' = Height between the parallel sides

From the given triangle ABE,

m(∠ABE) = m(∠ABC) - m(∠EBC)

m(∠ABE) = 120° - 90°

= 30°

By applying cosine rule in the given triangle,

cos(30°) = $\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

$\frac{\sqrt{3} }{2}=\frac{BE}{AB}$

$\frac{\sqrt{3} }{2}=\frac{BE}{6}$

BE = $3\sqrt{3}$ units

By applying sine rule in ΔABE,

sin(30°) = $\frac{\text{Opposite side}}{\text{Hypotenuse}}$

$\frac{1}{2}=\frac{AE}{AB}$

$\frac{1}{2}=\frac{AE}{6}$

AE = 3 units

Length of $b_1=BC=10$

Length of $b_2=AD=(AE+EF+FD)$ [AE = FD, since given trapezoid ABCD is an isosceles trapezoid]

$b_2=3+10+3$

$b_2=16$

Height between the parallel sides $h=3\sqrt{3}$

Area of the trapezoid = $\frac{1}{2}(BC+AD)BE$

= $\frac{1}{2}(10+16)(3\sqrt{3})$

= $39\sqrt{3}$

= 67.6 square units

8 0

3 years ago