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azamat
3 years ago
13

Which of the following figures could be classified as a rhombus?

Mathematics
1 answer:
pantera1 [17]3 years ago
3 0

Answer:

prolly say 5 ............

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Help im just doing this for my brother :C
cluponka [151]

Answer:

578

Step-by-step explanation:

118 + 560 = 678

678 - 100 = 578

578 is your answer

5 0
3 years ago
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Help me with geometry pleasee ( Radom answers will be reported )
IrinaVladis [17]
A. jgb=jb+Eh/2
35+45=80/2=40 JGB =40°
B. since it is a line HGJ =180-40
HGJ= 140°
C. tangents to circles have equal lengths if they intercept the way gd and gf do eachother
set the lengths equal to themselves
4x+5=6x-8
-4x both sides
5=2x-8
+8 both sides
13=2x
÷2 both sides
x=13/2 or 6.5
Plug in
4 (6.5)+5=
26+5=31ft
8 0
3 years ago
4/7<br> of a number is 36<br> What is the number?
katrin [286]

\text{If the number is x,}\\\\~~~~~\dfrac 47 \cdot x =36\\\\\implies 4x = 36 \times 7\\\\\implies x =\dfrac{36 \times 7}4\\ \\\implies x = 9 \times 7\\\\\implies x = 63\\\\\text{Hence, the number is 63.}

7 0
1 year ago
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2 to the power of 15<br> Is it correct to say like this?
vodka [1.7K]

Answer:

yes you can.

Step-by-step explanation:

...

7 0
2 years ago
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A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 31 ft/s. (a) A
julsineya [31]

Answer:

a) -13.9 ft/s

b) 13.9 ft/s

Step-by-step explanation:

a) The rate of his distance from the second base when he is halfway to first base can be found by differentiating the following Pythagorean theorem equation respect t:

D^{2} = (90 - x)^{2} + 90^{2}   (1)

\frac{d(D^{2})}{dt} = \frac{d(90 - x)^{2} + 90^{2})}{dt}

2D\frac{d(D)}{dt} = \frac{d((90 - x)^{2})}{dt}  

D\frac{d(D)}{dt} = -(90 - x) \frac{dx}{dt}   (2)

Since:

D = \sqrt{(90 -x)^{2} + 90^{2}}

When x = 45 (the batter is halfway to first base), D is:

D = \sqrt{(90 - 45)^{2} + 90^{2}} = 100. 62

Now, by introducing D = 100.62, x = 45 and dx/dt = 31 into equation (2) we have:

100.62 \frac{d(D)}{dt} = -(90 - 45)*31          

\frac{d(D)}{dt} = -\frac{(90 - 45)*31}{100.62} = -13.9 ft/s

Hence, the rate of his distance from second base decreasing when he is halfway to first base is -13.9 ft/s.

b) The rate of his distance from third base increasing at the same moment is given by differentiating the folowing Pythagorean theorem equation respect t:

D^{2} = 90^{2} + x^{2}  

\frac{d(D^{2})}{dt} = \frac{d(90^{2} + x^{2})}{dt}

D\frac{dD}{dt} = x\frac{dx}{dt}   (3)

We have that D is:

D = \sqrt{x^{2} + 90^{2}} = \sqrt{(45)^{2} + 90^{2}} = 100.63

By entering x = 45, dx/dt = 31 and D = 100.63 into equation (3) we have:

\frac{dD}{dt} = \frac{45*31}{100.63} = 13.9 ft/s

Therefore, the rate of the batter when he is from third base increasing at the same moment is 13.9 ft/s.

I hope it helps you!

4 0
3 years ago
Read 2 more answers
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