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34kurt
3 years ago
7

What ratio is equivalent to 80:88

Mathematics
2 answers:
Maksim231197 [3]3 years ago
6 0

Answer:

40:44

20:22

10:11

Step-by-step explanation:

bazaltina [42]3 years ago
3 0

Answer:

5040 : 5544

Step-by-step explanation:

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If sine theta equals three over four, what are the values of cos θ and tan θ?
ratelena [41]

Answer:

In words, Cosine theta equals plus or minus square root of seven over 4,tangent theta equals plus or minus  three over root seven

Step-by-step explanation:

Given that sin ∅ =3/4 It means the ratio of the opposite side to the hypotenuse side is 3:4.

Using the Pythagoras theorem we can calculate the hypotenuse adjacent as follows.

a²+b²=c²

a²=c²-b²

a²=4²-3²

a²=16-9

a²=7

a=√7

Then Cos ∅= opposite/ adjacent

=√7/4

Then Tan ∅ = opposite/adjacent

=3/√7

In words, Cosine theta equals plus or minus square root of seven over 4,tangent theta equals plus or minus  three over root seven.

7 0
3 years ago
Help help please please
gregori [183]

Answer:

c=√74

Step-by-step explanation:

h=c

p=5

b=7

Using pythogorean theorem,

h^2=p^2+b^2

c=√(7^2+5^2)

c=√(49+25)

c=√74

3 0
3 years ago
Read 2 more answers
In the equation above,a,b, and c are constants. If the equation is true for all values of x, what is the value of b?
DIA [1.3K]
Hope this answer helps.

6 0
3 years ago
Read 2 more answers
Which type of transformation is shown in the graph
Liono4ka [1.6K]

Answer:

C. Translation

Step-by-step explanation:

This is because the triangle is still in the same position, however, it has been shifted to the right five and up two.

Hope this helps!

All the love, Ya boi Fraser :)

4 0
3 years ago
Read 2 more answers
Verify:<br>cos(2A)=(cotA-tanA)/cscAsecA​
kolbaska11 [484]

Answer:

see explanation

Step-by-step explanation:

Using the trigonometric identities

cot A = \frac{cosA}{sinA}, tanA = \frac{sinA}{cosA}, cscA  = \frac{1}{sinA}, secA = \frac{1}{cosA}

Consider the right side

\frac{cotA-tanA}{cscAsecA}

= \frac{\frac{cosA}{sinA}-\frac{sinA}{cosA}  }{\frac{1}{sinA}.\frac{1}{cosA}  }

= \frac{\frac{cos^2A-sin^2A}{sinAcosA} }{\frac{1}{sinAcosA} }

= \frac{cos^2A-sin^2A}{sinAcosA} × sinAcosA ( cancel sinAcosA )

= cos²A - sin²A

= cos2A

= left side ⇒ verified

5 0
3 years ago
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