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2 years ago

Hello! Can anyone please tell me the measurement of t and v ? Any help is gladly appreciated.

Mathematics

1 answer:

makkiz [27]2 years ago

6 0

Answer:

v= 80, t=100

Step-by-step explanation:

yea

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Kitty [74]

In the ones place

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3 years ago

Read 2 more answers

Solve this equation using elimination: 4x-y=8;6x+y=2

fomenos

Solve simultaneously.

4x - y = 8
6x + y = 2

Multiply top equation by 6, and bottom equation by 4, to eliminate x, so that we can find y.
~
24x - 6y = 48
24x - 4y = 8

Subtract top from bottom to form one equation.
~
-2y = 40
Therefore y is 20.

Put y back in to an equation, such as 4x - y = 8.
~
4x - 20 = 8
4x = 28
x = 7

7 0

3 years ago

Find all missing angles

AleksandrR [38]

Answer:

a = 145

b = 35

c = 145

d = 70

e = 70

f = 110

g = 55

h = 125

i = 55

j = 50

k = 70

l =110

m = 50

n = 60

o = 70

p = 23

q = 89

r = 68

s = 157

t = 112

u = 48

v = 132

w = 132

x = 48

y = 48

z = 132

A = 48

B = 94

C = 86

D = 94

E = 47

F = 133

Step-by-step explanation:

I'm fairly certain these are all correct... like 82% sure

3 0

2 years ago

Mr Russo made a pizza that was 2 times as wide as a small pizza. His extra large pizza had a diameter of 18 inches. What was the

Pavlova-9 [17]

Diameter of small pizza = diameter of large pizza divided by 2 = 9 inches

x = y divided by 2
x = 9 inches

5 0

2 years ago

Solve the triangle A = 2 B = 9 C =8

VARVARA [1.3K]

Answer:

$\begin{gathered} A=\text{ 12}\degree \\ B=\text{ 114}\degree \\ C=54\degree \end{gathered}$

Step-by-step explanation:

To calculate the angles of the given triangle, we can use the law of cosines:

$\begin{gathered} \cos (C)=\frac{a^2+b^2-c^2}{2ab} \\ \cos (A)=\frac{b^2+c^2-a^2}{2bc} \\ \cos (B)=\frac{c^2+a^2-b^2}{2ca} \end{gathered}$

Then, given the sides a=2, b=9, and c=8.

$\begin{gathered} \cos (A)=\frac{9^2+8^2-2^2}{2\cdot9\cdot8} \\ \cos (A)=\frac{141}{144} \\ A=\cos ^{-1}(\frac{141}{144}) \\ A=11.7 \\ \text{ Rounding to the nearest degree:} \\ A=12º \end{gathered}$

For B:

$\begin{gathered} \cos (B)=\frac{8^2+2^2-9^2}{2\cdot8\cdot2} \\ \cos (B)=\frac{13}{32} \\ B=\cos ^{-1}(\frac{13}{32}) \\ B=113.9\degree \\ \text{Rounding:} \\ B=114\degree \end{gathered}$ $\begin{gathered} \cos (C)=\frac{2^2+9^2-8^2}{2\cdot2\cdot9} \\ \cos (C)=\frac{21}{36} \\ C=\cos ^{-1}(\frac{21}{36}) \\ C=54.3 \\ \text{Rounding:} \\ C=\text{ 54}\degree \end{gathered}$

3 0

1 year ago

Hello! Can anyone please tell me the measurement of t and v ? Any help is gladly appreciated.​

Hello! Can anyone please tell me the measurement of t and v ? Any help is gladly appreciated.