All categories

3 years ago

How many 5/6 in 10? Use models

1 answer:

4 0

10÷5/6
When dividing a number by a fraction, it is the same as multiplying by its reciprocal (the fraction flipped over)
so..
10÷5/6= 10×6/5= 60÷5= 12

You might be interested in

Nick found the quotient of 8.64 and 1.25 × 105. His work is shown below. 1. (8.64 × 101) (1.25 × 105) 2. (8.64 1.25 ) (101 105 )

Savatey [412]

Answer:

B.) No, the power multiplied to 8.64 should have an exponent of 0.

Step-by-step explanation:

took the test on Edgenuity and got it right! Hope this helps and please mark brainliest if can.

4 0

3 years ago

Read 2 more answers

Convert 20 mm into km

andrew-mc [135]

Answer:

Step-by-step explanation:

1 km = 1,000 m = 100,000 cm = 1,000,000 mm

1 mm is millionth part of 1 km

1 mm = 0.000001 km

20 mm = 0.00002 km

3 0

3 years ago

The Area of this triangle is 200ft ^2. What does x have to be.

Harlamova29_29 [7]

X=2.
Area for a triangle is .5(bh)
.5(5(2))*12=200.

8 0

3 years ago

Read 2 more answers

Urgent answer please!!! :)

Zepler [3.9K]

Answer:

a and b are relative on maxima interval x =-4,x=0

5 0

3 years ago

How to solve this derivative. f(x)={ln(3x)}^2x with steps...

AysviL [449]

Rewrite f(x) as a nested exponential-logarithm expression :

$\left(\ln(3x)\right)^{2x} = \exp\left(\ln\left(\ln(3x)\right)^{2x}\right)$

(where $\exp(x) = e^x$ )

One of the properties of logarithms lets us drop the exponent as a coefficient:

$\exp\left(\ln\left(\ln(3x)\right)^{2x}\right) = \exp\left(2x\ln\left(\ln(3x)\right)\right)$

Now, by the chain rule, we have

$f(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \implies \\\\ f'(x) = \left(\exp\left(2x\ln\left(\ln(3x)\right)\right)\right)' \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2x\ln\left(\ln(3x)\right)\right)'$

By the product rule,

$f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left( (2x)' \ln\left(\ln(3x)\right) + 2x\left(\ln\left(\ln(3x)\right)\right)' \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left( 2\ln\left(\ln(3x)\right) + 2x\left(\ln\left(\ln(3x)\right)\right)' \right)$

By the chain rule again,

$f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + 2x \cdot \dfrac1{\ln(3x)} \cdot \left(\ln(3x)\right)' \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2x}{\ln(3x)} \cdot \dfrac1{3x} \cdot (3x)' \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{3\ln(3x)} \cdot 3 \right) \\\\ f'(x) = \exp\left(2x\ln\left(\ln(3x)\right)\right) \cdot \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{\ln(3x)} \right)$

Then simplify this to

$f'(x) = \boxed{\left(\ln(3x)\right)^{2x} \left(2\ln\left(\ln(3x)\right) + \dfrac{2}{\ln(3x)} \right)}$

8 0

3 years ago