NEED HELP HW PLZ I WILL GIVe BRAINLIEST

What is the answer help

oksano4ka [1.4K]

Answer:

Step-by-step explanation:

4 0

3 years ago

Plz help!!!!!!!!!!!! Will be marked BRAINLIEST

s344n2d4d5 [400]

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bye

5 0

3 years ago

Read 2 more answers

A navy submarine descended at a rate of 4 feet per second. a. Write the descent rate as an integer. b. Write an expression to fi

Rasek [7]

Answer:

1200

Step-by-step explanation:

Since the value they are asking for is in minutes, we need to convert seconds to minutes. We can do that by multiplying 4 and 60, because there are 60 seconds in a minute, and we need to find the value of one minute. Also because the value they gave is 4. When we do this, we get 240. To get the final answer, we need to multiply 240 and 5, because 240 is the descending rate per minute, and they are asking for 5 minutes. When we do this, we get 1200.

(I might get this wrong, so like yeah.)

5 0

3 years ago

If A(4 -6) B(3 -2) and C (5 2) are the vertices of a triangle ABC fine the length of the median AD from A to BC. Also verify tha

Gnoma [55]

Answer:

a) The median AD from A to BC has a length of 6.

b) Areas of triangles ABD and ACD are the same.

Step-by-step explanation:

a) A median is a line that begin in a vertix and end at a midpoint of a side opposite to vertix. As first step the location of the point is determined:

$D (x,y) = \left(\frac{x_{B}+x_{C}}{2},\frac{y_{B}+y_{C}}{2} \right)$

$D(x,y) = \left(\frac{3 + 5}{2},\frac{-2 + 2}{2} \right)$

$D(x,y) = (4,0)$

The length of the median AD is calculated by the Pythagorean Theorem:

$AD = \sqrt{(x_{D}-x_{A})^{2}+ (y_{D}-y_{A})^{2}}$

$AD = \sqrt{(4-4)^{2}+[0-(-6)]^{2}}$

$AD = 6$

The median AD from A to BC has a length of 6.

b) In order to compare both areas, all lengths must be found with the help of Pythagorean Theorem:

$AB = \sqrt{(x_{B}-x_{A})^{2}+ (y_{B}-y_{A})^{2}}$

$AB = \sqrt{(3-4)^{2}+[-2-(-6)]^{2}}$

$AB \approx 4.123$

$AC = \sqrt{(x_{C}-x_{A})^{2}+ (y_{C}-y_{A})^{2}}$

$AC = \sqrt{(5-4)^{2}+[2-(-6)]^{2}}$

$AC \approx 4.123$

$BC = \sqrt{(x_{C}-x_{B})^{2}+ (y_{C}-y_{B})^{2}}$

$BC = \sqrt{(5-3)^{2}+[2-(-2)]^{2}}$

$BC \approx 4.472$

$BD = CD = \frac{1}{2}\cdot BC$ (by the definition of median)

$BD = CD = \frac{1}{2} \cdot (4.472)$

$BD = CD = 2.236$

$AD = 6$

The area of any triangle can be calculated in terms of their side length. Now, equations to determine the areas of triangles ABD and ACD are described below:

$A_{ABD} = \sqrt{s_{ABD}\cdot (s_{ABD}-AB)\cdot (s_{ABD}-BD)\cdot (s_{ABD}-AD)}$ , where $s_{ABD} = \frac{AB+BD+AD}{2}$

$A_{ACD} = \sqrt{s_{ACD}\cdot (s_{ACD}-AC)\cdot (s_{ACD}-CD)\cdot (s_{ACD}-AD)}$ , where $s_{ACD} = \frac{AC+CD+AD}{2}$

Finally,

$s_{ABD} = \frac{4.123+2.236+6}{2}$

$s_{ABD} = 6.180$

$A_{ABD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}$

$A_{ABD} \approx 3.004$

$s_{ACD} = \frac{4.123+2.236+6}{2}$

$s_{ACD} = 6.180$

$A_{ACD} = \sqrt{(6.180)\cdot (6.180-4.123)\cdot (6.180-2.236)\cdot (6.180-6)}$

$A_{ACD} \approx 3.004$

Therefore, areas of triangles ABD and ACD are the same.

4 0

4 years ago

To prepare for the town's race, Andrew runs around a rectangular field. The dimensions of the field are 450 feet

Ket [755]

Answer:

46.9 times

Step-by-step explanation:

First, we can calculate how many feet it takes to run around the field. To find the perimeter of a rectangle, we can use the formula

2* length + 2 * width. With the length being 450 and the width being 225 here, we can say that

2*450 + 2 * 225 = 1350 feet

Therefore, Andrew runs 1350 feet each time he runs around the field. Next, we need to figure out how much 1350 feet goes into 12 miles as we want to find how many times Andrew runs around the field to get to 12 miles. This can be represented by

12 miles/1350 feet

One thing that we can do here is multiply the fraction by 1 to keep it the same. Because 1 mile = 5280 feet, we can say that

1 mile/5280 feet = 1 = 5280 feet/1 mile. Therefore, it would be safe to multiply

12 miles/1350 feet by 1 = 5280 feet/1 mile. Note that feet is on the bottom in the first fraction (12 miles/1350 feet) and on the top in the second (5280 feet/1 mile) so they will cancel out. Similarly, miles are on top in the first and bottom in the second. We then have

12 miles/1350 feet * 5280 feet/1 mile =63360/1350 ≈ 46.9

4 0

3 years ago