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3 years ago

The composite figure is made up of three simpler shapes? what is the area of the figure

Mathematics

1 answer:

worty [1.4K]3 years ago

8 0

Answer:

B. 68 square cm

Step-by-step explanation:

0.5 x 6 x 5 = 15

5 x 7 = 35

9 x 2 = 18

15+35+18 =68

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Help me out please?<br>chapter : exponents and powers

belka [17]

Answer:

$\frac{125}{27}$

Step-by-step explanation:

$\frac{2^{-4} \times 15^{-3} \times 625}{5^2 \times 10^{-4}}$

Lets expand all the composite numbers into prime numbers.

$=> \frac{2^{-4} \times (3^{-3} \times 5^{-3}) \times 5^4}{5^2 \times (2^{-4} \times 5^{-4})}$

Lets cancel $2^{-4}$ from numerator and denominator.

$=> \frac{3^{-3} \times 5^{-3} \times 5^4}{5^2 \times 5^{-4}}$

Using laws of exponents , lets solve this.

$=> \frac{3^{-3} \times 5^{(-3 + 4)}}{5^{( 2 - 4)}}$

$=> \frac{3^{-3} \times 5^{1}}{5^{-2}}$

$=> 3^{-3} \times 5^{[1 - (-2)]}$

$=> 3^{-3} \times 5^{3}$

$=> \frac{5^3}{3^3} = \frac{125}{27}$

4 0

2 years ago

Read 2 more answers

In 1 hour Samantha can make 6 beaded bracelets. In 3 hours she can make 18 bracelets. How many can she make in 4 fours? PLEASE P

Step2247 [10]

Basically Samantha makes six bracelets per hour, so:
After 1 hour she makes 6
After 2 hours she makes 12
After 3 hours she makes 18
After 4 hours she makes 24, therefor the answer is B

This can also be modelled by the equation B = 6t, where B is the number of bracelets and t is the number of hours, so:
B = 6(4) = 24

3 0

3 years ago

You are a lifeguard and spot a drowning child 60 meters along the shore and 40 meters from the shore to the child. You run along

sukhopar [10]

Answer:

The lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.

Step-by-step explanation:

This is a problem of optimization.

We have to minimize the time it takes for the lifeguard to reach the child.

The time can be calculated by dividing the distance by the speed for each section.

The distance in the shore and in the water depends on when the lifeguard gets in the water. We use the variable x to model this, as seen in the picture attached.

Then, the distance in the shore is d_b=x and the distance swimming can be calculated using the Pithagorean theorem:

$d_s^2=(60-x)^2+40^2=60^2-120x+x^2+40^2=x^2-120x+5200\\\\d_s=\sqrt{x^2-120x+5200}$

Then, the time (speed divided by distance) is:

$t=d_b/v_b+d_s/v_s\\\\t=x/4+\sqrt{x^2-120x+5200}/1.1$

To optimize this function we have to derive and equal to zero:

$\dfrac{dt}{dx}=\dfrac{1}{4}+\dfrac{1}{1.1}(\dfrac{1}{2})\dfrac{2x-120}{\sqrt{x^2-120x+5200}} \\\\\\\dfrac{dt}{dx}=\dfrac{1}{4} +\dfrac{1}{1.1} \dfrac{x-60}{\sqrt{x^2-120x+5200}} =0\\\\\\ \dfrac{x-60}{\sqrt{x^2-120x+5200}} =\dfrac{1.1}{4}=\dfrac{2}{7}\\\\\\ x-60=\dfrac{2}{7}\sqrt{x^2-120x+5200}\\\\\\(x-60)^2=\dfrac{2^2}{7^2}(x^2-120x+5200)\\\\\\(x-60)^2=\dfrac{4}{49}[(x-60)^2+40^2]\\\\\\(1-4/49)(x-60)^2=4*40^2/49=6400/49\\\\(45/49)(x-60)^2=6400/49\\\\45(x-60)^2=6400\\\\$

$x$

As $d_b=x$ , the lifeguard should run across the shore a distance of 48.074 m before jumpng into the water in order to minimize the time to reach the child.