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Naya [18.7K]
3 years ago
13

Law of cosines: a2 = b2 + c2 – 2bccos(A)

Mathematics
2 answers:
andrezito [222]3 years ago
8 0

The correct answer is:

D. y^{2}=9^{2}+19^{2}-2(9)(19)cos(41)

|Huntrw6|

damaskus [11]3 years ago
3 0
In this case,
a = y
b = 9
c = 19
So...
a^2 = b^2 + c^2 - 2bc*\cos(A)
y^2 = 9^2 + 19^2 - 2*9*19*\cos(41^{\circ})
which means the answer is choice D
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Which quadratic function has a wider graph than y=2x^2?
Trava [24]

Answer:

y = 1/2 x²

Step-by-step explanation:

The coefficient of the first term in a quadratic, in our case here, x², will tell us how the graph stretches. This is akin to the slope within the linear graph. Similar to the slope, the smaller the coefficient value, or value of slope m, the shallower the angle.

When discussing quadratics, the larger the coefficient of our x² term, the steeper, and skinnier the graph. If we want to look for a graph that is wider than y = 2x², then we need to find a graph with a coefficient that is less than 2.

Our only option then is

y = 1/2 x²

8 0
3 years ago
Find the unknow side of the right triangle 2. b=73.6 cm b= 19.0
shusha [124]
Assumed 73.6cm is the hypotenuse.

a^{2} + b^{2} = c^{2}
19^{2} + b^{2} = 73.6^{2}
b^{2} = 73.6^{2} -19^{2}
b^{2} = 5055.96
b = 71.1

Assumed 73.6cm is not the hypotenuse.
a^{2} + b^{2} = c^{2}
19^{2} + 73.6^{2} = c^{2}
c^{2} = 73.6^{2} +19^{2}
c^{2} = 5777.96
c = 76.0


7 0
3 years ago
The number of loaves of bread purchased and the total cost of the bread in dollars can be modeled by the equation c = 3.5b. Whic
Valentin [98]

Answer:

Loaves of bread: b can't be a negative value, then the first table doesn't include viable solutions.

The number of loaves of bread must be a whole number, then the second table doesn't include viable solutions.

c=3.5b

if b=0→c=3.5(0)→c=0

If b=3→c=3.5(3)→c=10.5

If b=6→c=3.5(6)→c=21

If b=9→c=3.5(9)→c=31.5

The third table includes viable solutions

4 0
3 years ago
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Minchanka [31]

whats that game about

8 0
3 years ago
A) If a:b = 2:5 and b:c = 3:4, find (i) a:c(ii) a:b:c
kobusy [5.1K]

Answer:

A i. a:c=3:10

ii. a:b:c=2:5:10

B i. x:z=2:5

ii. x:y:z=2:4:5

Step-by-step explanation:

A.) If a:b = 2:5 and b:c = 3:4, find (i) a:c(ii) a:b:c

a:b=a/b=2/5

b:c=b/c=3/4

a/b*b/c=a/c

2/5*3/4=a/c

6/20=a/c

3/10=a/c

Therefore, a:c=3:10

a:b:c

a:b=2:5

b:c=3:4

b is common to both ratios

The value of b in the first ratio is 5 and b is 3 in the second ratio

Lets take the LCM of both values

LCM of 5 and 3=15

So, we will change the value of b in the first ratio and second ratio to 15

By doing this, we will multiply the whole first ratio by 3

We have, 6:15

We multiply the whole second ratio by 5

We have, 15:20

Therefore a:b:c=6:15:20

=2:5:10

B. If x:y = 1:2 and y:z = 4:5,

x:y=x/y=1:2

y:z=y/z=4:5

x/y*y/z=x/z

1/2*4/5=x/z

4/10=x/z

2/5=x/z

Therefore, x:z=2:5

x:y:z

x:y=1:2

y:z=4:5

y is common to both ratio

Take the LCM of y values in both ratio

LCM of 2 and 4 =4

So,we will change the value of y in the first and second ratio to 4

By doing this, we will multiply the whole first ratio by 2

We have, 2:4

We will also multiply the whole second ratio by 1

We have, 4:5

Therefore, x:y:z=2:4:5

8 0
3 years ago
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