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3 years ago

A school held a jump-roping contest. Diego jumped rope for 20 minutes.

Mathematics

1 answer:

professor190 [17]3 years ago

5 0

Answer:

Jada jumped 75% of what Diego jumped.

Step-by-step explanation:

75% of 20 is 15. So it is 75%.

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1)a chord of length 18cm midway the radius of a circle. calculate the radius of the circle correct to 1d.p. 2)if two parallel ch

kykrilka [37]

Part 1:

Given that the length of the chord is 18 cm and the chord is midway the radius of the circle.

Thus, half the angle formed by the chord at the centre of the circle is given by:

$\cos\theta=\frac{\left( \frac{1}{2} r\right)}{r}= \frac{1}{2} \\ \\ \Rightarrow\theta=\cos^{-1}\left( \frac{1}{2} \right)=60^o$

Now,

$\sin60^o= \frac{9}{r} \\ \\ \Rightarrow r= \frac{9}{\sin60^o} =10.392$

Therefore, the radius of the circle is 10.4 cm to 1 d.p.

Part 2I:

Given that the radius of the circle is 10 cm and the length of chord AB is 8 cm. Thus, half the length of the chord is 4cm. Let the distance of the mid-point O to /AB/ be x and half the angle formed by the chord at the centre of the circle be θ, then

$\sin\theta= \frac{4}{10} = \frac{2}{5} \\ \\ \theta=\sin^{-1}\left( \frac{2}{5} \right)=23.6^o$

Now,

$\cos23.6^o= \frac{x}{10} \\ \\ \Rightarrow x=10\cos23.6^o=9.165\approx9.2cm$

Part 2II:

Given that the radius of the circle is 10cm and the angle distended is 80 degrees. Let half the length of chord CD be y, then:

$\sin40^o= \frac{y}{10} \\ \\ \\ \Rightarrow y=10\sin40^o=6.428$

Thus, the length of chord CD = 2(6.428) = 12.856 which is approximately 12.9 cm.

3 0

3 years ago

An art teacher gives a total of 25 pounds of clay to her students. She gives each of her 12 students the same amount of clay. Ho

Alenkasestr [34]

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5 0

3 years ago

Read 2 more answers

Which of the following are not congruence theorems for right triangles? Check all that apply.

Lera25 [3.4K]

Answer:

c) HH E)AA

Step-by-step explanation:

Two right angles can be proved congruent by LA,LL,HA,HL properties.

HH and AA are not congruence property.

Two triangles can be proved similar by AA property. All congruent triangles are similar but all similar triangles are not similar.Hence right triangles can not be proved congruent by AA property.

Among the options given Option C and E are not congruence property.

7 0

3 years ago

Read 2 more answers

For which of the following equations are x = 5 and x = –5 both solutions?

Minchanka [31]

Short Answer B
Argument
A
A will give you x = +/- 5i
x^2 + 25 = 0
x^2 = - 25 Take the square root.
sqrt(x^2) = +/- sqrt(-25)
x = +/- (5)i which is a complex number.

B
Is the answer
x^2 = 25
sqrt(x)^2 = sqrt(25)
x = +/- 5

C
Can't be factored just by looking at it. You can show that C is not true just by putting 5 into the equation
f(x) = x^2 + 10x - 25
f(5) = 25 + 10*5 - 25
f(5) = 50
C is not true.

D
D can be eliminated as C was
f(x) = x^2 - 5x - 25
f(5) = -25 ( l'll let you show this is not true). 5 is not a solution because it does not make f(x) = 0

6 0

3 years ago

Read 2 more answers

Which inequality is true?

aleksandrvk [35]

Answer:

I don't know I don't know about question

3 0

3 years ago