All categories

3 years ago

Helppppppppp i will give points

Mathematics

1 answer:

seraphim [82]3 years ago

7 0

Pretty sure that would be 120. Since the angles of a triangle add up to 180, you just add 25 and 35 (makes 60) and subtract it by 180, leaving 120

You might be interested in

Evaluate,

kirill [66]

With 10 to the power to a negative number, it's easier.
Here's a trick: The negative number is the places to the right of the decimal. The last number is a 1, while the rest are 0s.

From this, we get: $10^{-4}=\boxed{0.0001}$

5 0

3 years ago

Read 2 more answers

Mr. Morris is going to save money and replace his sailboat's mainsail himself. He must determine the area of the mainsail in ord

e-lub [12.9K]

The decomposition of the parallelogram into triangle then into a rectangle is shown below

The measurement we need to calculate the area is the length and the width of the rectangle; 15 feet and 10 feet

Area = 15 × 10 = 150 feet²

4 0

3 years ago

Help soon eeeeeeeeeeeeeeeeeee

igomit [66]

3/9 because it is the same as 1/3 or 2/6.

6 0

3 years ago

Read 2 more answers

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.

7 0

3 years ago

What grade level is a score of 235 in maps math I am really wanting to know

igomit [66]

Answer: It should be an 8th grade level

(sorry if i'm wrong)

6 0

3 years ago