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AlladinOne [14]
3 years ago
14

In the diagram, JK || GI. If the ratio of HJ to JG is 3 to 1, then what is the ratio of HK to KI?

Mathematics
1 answer:
Ket [755]3 years ago
5 0
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Solve for m: E=mc²<br> please help
aleksandr82 [10.1K]

Step-by-step explanation:

we have,

E = mc2

or, E/c2 = m

here is your answer..

3 0
3 years ago
Find the following, giving your answers to 1 decimal place. 10 cm 8 cm 6 cm a) The volume of the cone. b) The curved surface are
sesenic [268]

Answer:

Volume of cone = 209.44 cm²

Curved surface area of cone = 188.57 cm

Step-by-step explanation:

Given:

Height h = 8 cm

Slant height l = 10 cm

Radius r = 6 cm

Find:

Volume of cone

Curved surface area of cone

Computation:

Volume of cone = (1/3)πr²h

Volume of cone = (1/3)π(5)²(8)

Volume of cone = 209.44 cm²

Curved surface area of cone = πrl

Curved surface area of cone = π(6)(10)

Curved surface area of cone = 188.57 cm

3 0
3 years ago
Find two angles (in radians) whose sine ratio is...
frutty [35]

Hi there!

\large\boxed{\frac{\pi}{3} \text{ and } \frac{2\pi}{3}}

From the unit circle, we can locate the two values of sine Ф (y-value of the coordinate)  at which sin Ф = √3/2.

The two values are Ф = π/3 and 2π/3.

8 0
3 years ago
F(x)=3x^2-18x+27<br><br>can you solve this?
Alika [10]
What do you mean by "solve this?"  Until you set f(x) = to some constant number, you don't have an equation and thus can't expect to find solutions.

Why don't we take <span>3x^2-18x+27 and set it = to 0, and only then try to solve this equation?

</span><span>3x^2-18x+27 = 0.  Dividing both sides by 3, we get x^2 - 6x + 9= 0, which can be factored as

(x-3)^2 = 0.  Taking the sqrt of both sides, we get x-3 = 0.  Actually, there are 2 roots:  3 and 3.  Again, this statement is true only if we set </span><span>3x^2-18x+27 = to 0.</span>
6 0
3 years ago
Read 2 more answers
Calculate an estimate of a square root of 119 + 120i
uysha [10]

Answer:

Step-by-step explanation:

Given

z=119+120 i

Let \sqrt{119+120 i}=p+iq

Squaring both sides

119+120 i=p^2-q^2+2ipq

Comparing real and imaginary part

Re(LHS)=Re(RHS)

119=p^2-q^2-----------1

comparing Im(LHS)=Im(RHS)

120=2pq

q=\frac{60}{p}

Substitute q in 1

119=p^2-(\frac{60}{p})^2

p^4-119p^2-(68)^2=0

Let x=p^2

x^2-119x-4624=0

x=frac{119\pm \sqrt{119^2+4\times 4624}}{2}

x=\frac{119\pm 180.71}{2}

we take only Positive value because p^2=x

x=149.85

p^2=149.85

thus p=\pm 12.24

q=\mp 4.90

thus \sqrt{119+120 i}=\pm (12.24+i 4.90)

6 0
4 years ago
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