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ICE Princess25 [194]
3 years ago
10

HELP ME PLEASE

Mathematics
1 answer:
dsp733 years ago
3 0
Preyton took longer so he was slower. Now, we calculate the average speeds of both:
Preyton: 150 / 3 = 50 mph
Bryson: 150 / 2.5. = 60 mph

Therefore, Bryson travelled 10 mph faster than Preyton
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A fence is 10 feet tall. A stick leans against the fence as shown. What is the length of the stick?
Vitek1552 [10]
The answer should be 22 1/2 ft
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Encik Basra’s age is four times his child’s age. 4 years ago, Encik Basri’s age is six times his child age. Find the age of Enci
Luden [163]

Answer:

4 + 6 = 10 * 4 =40

Step-by-step explanation:

sana makatulong

4 0
3 years ago
Suppose you are the diving officer on a submarine conducting diving operations. As you conduct your operations, you realize that
madam [21]

Answer:

If you need additional resources to support virtual learning, please ... 9) I can solve problems involving a system of linear inequalities. ... of the submarine is 50 ft below sea level when it starts to descend ... ft/s. It dives at that rate for 5 s. ... After the initial 5-second descent, the submarine increases its rate of ...

Step-by-step explanation:

4 0
3 years ago
Simplify : (√ + √ ) (√ − √) ( + )( 2 + 2 )
jek_recluse [69]

(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})(x+y)(x^2+y^2)\\\\=[(\sqrt{x})^2-(\sqrt{y})^2](x+y)(x^2+y^2)\\\\=(x-y)(x+y)(x^2+y^2)\\\\=(x^2-y^2)(x^2+y^2)\\\\=(x^2)^2-(y^2)^2\\\\=x^4-y^4


Used:\\\\(\sqrt{a})^2=a\\\\(a-b)(a+b)=a^2-b^2\\\\(a^n)^m=a^{nm}

6 0
3 years ago
What is the perimeter of the triangle shown on the coordinate plane,to the nearest tenth of a unit ?
ollegr [7]

Answer:

25.6 units

Step-by-step explanation:

From the figure we can infer that our triangle has vertices A = (-5, 4), B = (1, 4), and C = (3, -4).

First thing we are doing is find the lengths of AB, BC, and AC using the distance formula:

d=\sqrt{(x_2-x_1)^{2} +(y_2-y_1)^{2}}

where

(x_1,y_1) are the coordinates of the first point

(x_2,y_2) are the coordinates of the second point

- For AB:

d=\sqrt{[1-(-5)]^{2}+(4-4)^2}

d=\sqrt{(1+5)^{2}+(0)^2}

d=\sqrt{(6)^{2}}

d=6

- For BC:

d=\sqrt{(3-1)^{2} +(-4-4)^{2}}

d=\sqrt{(2)^{2} +(-8)^{2}}

d=\sqrt{4+64}

d=\sqrt{68}

d=8.24

- For AC:

d=\sqrt{[3-(-5)]^{2} +(-4-4)^{2}}

d=\sqrt{(3+5)^{2} +(-8)^{2}}

d=\sqrt{(8)^{2} +64}

d=\sqrt{64+64}

d=\sqrt{128}

d=11.31

Next, now that we have our lengths, we can add them to find the perimeter of our triangle:

p=AB+BC+AC

p=6+8.24+11.31

p=25.55

p=25.6

We can conclude that the perimeter of the triangle shown in the figure is 25.6 units.

6 0
3 years ago
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