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Soloha48 [4]
3 years ago
11

the ratio of clown fish to angel fish at a pet store is 5: 4. the ratio of angle fish to goldfish is4:3 there are 60 clown fish

at the pet store
Mathematics
1 answer:
denis23 [38]3 years ago
3 0
Clown fish: angel fish    5:4
angle fish: gold fish      4:3

(60): angel fish=5:4
(5)(12)=60
(4)(12)=48 angel fish

(48): gold fish=4:3
(4)(12)=48
(3)(12)=36 gold fish

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please give solutions kindly write and send pic in my whatsap no.( eight zero one seven zero zero zero nine five zero)​
zloy xaker [14]

Answer:

You simply take the values away from 180°.

a) 180-85-65 = 30°

b) 180-37-63=80°

c) 180-52-78=50°

d) 180-102-58=20°

e) 180-100-60=20°

f) 180-105-25=50°

g) 180-127-23=30°

6 0
2 years ago
Which steps can be used to solve<br> 6<br> X<br> 1 7<br> 2 8<br> for x?
Gnesinka [82]

Answer:

2856

Step-by-step explanation:

This question is a little unclear. 6 times 17 times 28 is equal to 2856 if that is what you are trying to ask.

4 0
2 years ago
Verify sine law by taking triangle in 4 quadrant<br>Explain with figure.<br>​
Ksivusya [100]

Proof of the Law of Sines

The Law of Sines states that for any triangle ABC, with sides a,b,c (see below)

a

 sin  A

=

b

 sin  B

=

c

 sin  C

For more see Law of Sines.

Acute triangles

Draw the altitude h from the vertex A of the triangle

From the definition of the sine function

 sin  B =

h

c

    a n d        sin  C =

h

b

or

h = c  sin  B     a n d       h = b  sin  C

Since they are both equal to h

c  sin  B = b  sin  C

Dividing through by sinB and then sinC

c

 sin  C

=

b

 sin  B

Repeat the above, this time with the altitude drawn from point B

Using a similar method it can be shown that in this case

c

 sin  C

=

a

 sin  A

Combining (4) and (5) :

a

 sin  A

=

b

 sin  B

=

c

 sin  C

- Q.E.D

Obtuse Triangles

The proof above requires that we draw two altitudes of the triangle. In the case of obtuse triangles, two of the altitudes are outside the triangle, so we need a slightly different proof. It uses one interior altitude as above, but also one exterior altitude.

First the interior altitude. This is the same as the proof for acute triangles above.

Draw the altitude h from the vertex A of the triangle

 sin  B =

h

c

      a n d          sin  C =

h

b

or

h = c  sin  B       a n d         h = b  sin  C

Since they are both equal to h

c  sin  B = b  sin  C

Dividing through by sinB and then sinC

c

 sin  C

=

b

 sin  B

Draw the second altitude h from B. This requires extending the side b:

The angles BAC and BAK are supplementary, so the sine of both are the same.

(see Supplementary angles trig identities)

Angle A is BAC, so

 sin  A =

h

c

or

h = c  sin  A

In the larger triangle CBK

 sin  C =

h

a

or

h = a  sin  C

From (6) and (7) since they are both equal to h

c  sin  A = a  sin  C

Dividing through by sinA then sinC:

a

 sin  A

=

c

 sin  C

Combining (4) and (9):

a

 sin  A

=

b

 sin  B

=

c

 sin  C

7 0
3 years ago
Identify the vertex of each graph. Tell whether it is a minimum or a maximum
suter [353]

Answer:

1,3 minimum

1,6 minimum

3,1 maximum

Step-by-step explanation:

Locate the h as x and the k as y for y-k=a(x-h)^2

4 0
3 years ago
Please help <br>Much appreciated ​
mafiozo [28]

Answer:

x^2 -10x+25

Step-by-step explanation:

(x-5)^2

Rewriting

(x-5)(x-5)

FOIL

first: x*x = x^2

outer: x* -5 = -5x

inner: -5*x = -5x

last: -5 * -5 = +25

Add them together

x^2 -5x-5x+25

Combine

x^2 -10x+25

8 0
3 years ago
Read 2 more answers
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