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4 years ago

What is 4 2/7 ⋅ 2 1/5 ?

Mathematics

2 answers:

Solnce55 [7]4 years ago

4 0

Answer:

The answer is 9 3/7

Step-by-step explanation:

Vikki [24]4 years ago

3 0

Answer: 9 and 3/7

Step-by-step explanation: To multiply a mixed number by a mixed number, we first want to write each mixed number as an improper fraction.

We can write a mixed number as an improper fraction by multiplying the denominator by the whole number and adding the numerator.

We can change $4\frac{2}{7}$ to the improper fraction 30/7.

We can also change $2\frac{1}{5}$ to the improper fraction 11/5.

Now, we are simply multiplying fractions. To multiply fractions, we multiply across the numerators and multiply across the denominators. When we multiply these fractions, we will end up with the fraction 330/35. Since 330/35 is not in lowest terms, we need to divide both the numerator and denominator by the greatest common factor of 330 and 35 which is 5 and we will get the improper fraction 66/7.

Notice however that 66/7 can be changed into a mixed number by dividing the denominator which is 7 into the numerator which is 66.

66 divided by 7 equals 9 with a remainder of 3 so we can write 66/7 as the mixed number 9 and 3/7.

Therefore, 4 and 2/7 × 2 and 1/5 = 9 and 3/7

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Find the solution of the initial value problem y Superscript prime prime Baseline plus 4 y Superscript prime Baseline plus 5 y e

sergiy2304 [10]

Answer:

$y(t) = - 5 {e}^{2\pi - 2t} \cos(t)$

Step-by-step explanation:

The given initial value problem is

$y''+4y'+5=0$

$y( \frac{ \pi}{2} ) = 0$

$y'( \frac{ \pi}{2} ) = 5$

The corresponding characteristic equation is

${m}^{2} + 4m + 5 = 0$

$m = - 2 \pm \: i$

The general solution becomes:

$y(t) = A {e}^{ - 2t} \cos(t) + B {e}^{ - 2t} \sin(t)$

We differentiate to get:

$y'(t) = - A {e}^{ - 2t} \sin(t) - 2 A {e}^{ - 2t} \cos(t) + B {e}^{ - 2t} \cos(t) - 2B {e}^{ - 2t} \sin(t)$

We apply the initial conditions to get;

$y( \frac{\pi}{2} ) = A {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) + B {e}^{ - 2\pi} \sin( \frac{\pi}{2} )$

$A {e}^{ - 2\pi} (0 ) + B {e}^{ - 2\pi} ( 1) = 0$

$B = 0$

Also;

$y'( \frac{\pi}{2} ) = - A {e}^{ - 2\pi} \sin( \frac{\pi}{2} ) - 2 A {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) + B {e}^{ - 2\pi} \cos( \frac{\pi}{2} ) - 2B {e}^{ - 2\pi} \sin( \frac{\pi}{2} )$