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3 years ago

Help meeee photo is attached below

Mathematics

1 answer:

Alex787 [66]3 years ago

6 0

Answer:

I don't see it I think you forgot to attach it

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HELP PLEASE

mario62 [17]

Answer:

26.4

Step-by-step explanation:

Law Of Cosines:

$cos(A)=\frac{b^2+c^2-a^2}{2bc}$

This should work for any side. This can generally be thought as:

$cos(\text{angle}) = \frac{\text{sum of squares of two other sides-opposite side squared}}{\text{2 times the product of the other two sides}}$

If this is too confusing here's the formula for the other sides (which is essentially the same, just different variables)

$cos(B)=\frac{a^2+c^2-b^2}{2ac}$

$cos(C) =\frac{a^2+b^2-c^2}{2ab}$

Anyways now just plug in the known values into the equation

$cos(A)=\frac{4^2+6^2-3^2}{2(6)(4)}\\$

Square and multiply values

$cos(A)=\frac{16+36-9}{48}$

Add the values in the numerator

$cos(A)=\frac{43}{48}$

Take the inverse of cosine on both sides

$A=cos^{-1}(\frac{43}{48})$

calculate arccosine (inverse cosine) using a calculator

$A\approx 26.384$

Round to nearest tenth

$A\approx26.4$

7 0

2 years ago

The picture above The picture above The picture above

Semenov [28]

Answer:

dfxgsdgsgagbffgdf

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

What is the measure of

gizmo_the_mogwai [7]

Answer: Q = 124 and P = 29

Step-by-step explanation:

Triangles always = 180 degrees so do straight lines.

6 0

3 years ago

Math: Please help me I don’t understand and I need to do corrections.

GREYUIT [131]

So, you had done everything right so far (other than squaring the 2), but that was only half of the question.

to find the least common multiple, you need to first figure out what the prime factors have in common.
${2}^{2} \times 3 \times 5 \\ and \\ {2}^{2} \times {3}^{2} \times 5 \times 7$
each have two twos. both have one 5, so we know our answer will look something like
${2}^{2} \times 5 \times other \: stuff$
now to figure out the other stuff... we have to represent the greatest amount of everything that is left, and we have 3s and 7s left over, so we need to figure out how many of each we need.

one has one 3 and one has two, so we need two threes. now our equation is
${2}^{2} \times {3}^{2} \times 5 \times stuff$

what's the only number we have to deal with? 7...

how many sevens does 60 have? 0, and 630 has 1, so we know we need one 7. our answer becomes

4 0

3 years ago

Solve the inequality. 2(4+2x)25x+5

Wewaii [24]

2(4+2x)25x+5 = 300x + 5
2(4+2)•25x+5
2(4+2)•25x=300x
= 300x + 5

6 0

3 years ago