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2 years ago

A = 16, b = 20, mZ B = 40 Law of cosines

Mathematics

1 answer:

bezimeni [28]2 years ago

4 0

What is this ?chfufuykygiufffjfufuf

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the area of a rectangular field 4641 m squared. If the width of the field is 51 m, what is its length?

shusha [124]

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

Hello!

✧･ﾟ: *✧･ﾟ:* *:･ﾟ✧*:･ﾟ✧

❖ The length of the field is 91 m.

Divide to find the missing length:

4641 ÷ 51 = 91

~ ʜᴏᴘᴇ ᴛʜɪꜱ ʜᴇʟᴘꜱ! :) ♡

~ ᴄʟᴏᴜᴛᴀɴꜱᴡᴇʀꜱ

5 0

2 years ago

Read 2 more answers

Please help i really don't understand

Natali5045456 [20]

Answer:

80 square units

Step-by-step explanation:

The area formula refers to a generic triangle ABC in which side lengths 'a' and 'b' are known and angle C is between those sides.

In the given figure, we have known side lengths of 12 and 14, and the angle between them is 72°.

Putting these numbers into the formula, we find the area to be ...

A = (1/2)(12)(14)sin(72°) ≈ 79.9 ≈ 80 . . . . square units

The area of the triangle is about 80 square units.

7 0

2 years ago

Find a · b. |a| = 80, |b| = 50, the angle between a and b is 3π/4.

mart [117]

Answer:

$a\cdot b=-2000\sqrt{2}$

Step-by-step explanation:

Given information: |a| = 80, |b| = 50, the angle between a and b is 3π/4.

We need to find the dot product a · b.

The formula of dot product is

$a\cdot b=|a||b|\cos \theta$

where, θ is the angle between a and b.

Substitute the given values in the above formula.

$a\cdot b=(80)(50)\cos (\frac{3\pi}{4})$

$a\cdot b=4000\cos (\pi-\frac{\pi}{4})$

$a\cdot b=-4000\cos (\frac{\pi}{4})$ $[\because \cos (\pi-\theta)=-\cos \theta]$

$a\cdot b=-4000\frac{1}{\sqrt{2}}$

$a\cdot b=-\frac{4000}{\sqrt{2}}$

Rationalize the above equation.

$a\cdot b=-\frac{4000}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$

$a\cdot b=-\frac{4000\sqrt{2}}{2}$

$a\cdot b=-2000\sqrt{2}$

Therefore, the value of a · b is $a\cdot b=-2000\sqrt{2}$ .

7 0

2 years ago

Read 2 more answers

Determine whether LINE AB and LINE CD are parallel, perpendicular, or neither. A(−1, −4) , B(2, 11) , C(1, 1) , D(4, 10) ( write

Luba_88 [7]

Step-by-step explanation:

Hey there!

The points of line AB are; (-1,-4) and (2,11).

Note:

Use double point formula and simplify it to get two eqaution.
Use condition of parallel lines, perpendicular lines to know whether the lines are parallel or perpendicular or nothing.

~ Use double point formula.

$(y - y1) = \frac{y2 - y1}{x2 - x1} (x - x1)$

~ Keep all values.

$(y + 4) = \frac{11 + 4}{2 + 1} (x + 1)$

~ Simplify it.

$y + 4 = \frac{15}{3} (x + 1)$

$y + 4 = 5x + 5$

$5x - y + 1 = 0$

Therefore this is the equation of line AB.

Now, Finding the equation of line CD.

Given;

The points of line CD are; (1,1) and (4,10).

~ Using formula.

$(y - y1) = \frac{y2 - y1}{x2 - x1}(x - x1)$

~ Keep all values.

$(y - 1) = \frac{10 - 1}{4 - 1} (x - 1)$

~ Simplify it.

$y - 1 = 3 x - 3$

$3x - y - 2 = 0$

Therefore, 3x - y- 2 = 0 is the eqaution of line CD.

Use condition of parallel lines.

m1= m2

Slope of equation (i)

$m1 = \frac{ - coeff. \: of \: x}{coeff \: of \: y}$

$m1 = \frac{ - 5}{ - 1}$

Therefore, m1 = 5

Slope of second equation.

$m2 = \frac{ - coeff \: .of \: x}{coeff \: .of \: y}$

$m2 = \frac{ - 3}{ - 1}$

Therefore, m2 = 3.

Now, m1≠m2.

So, the lies are not parallel.

Check for perpendicular.

m1*m2= -1

3*5≠-1.

Therefore, they aren't perpendicular too.

So, they are neither.

Hope it helps...

6 0

2 years ago

Use the substitution method to solve the system of equations. Choose the

Allisa [31]

$\bf \begin{cases} x+2y=12\\ \cline{1-1} -x=-y-6\\ \boxed{x}=y+6 \end{cases}\qquad \qquad \stackrel{\textit{doing some substitution in the 1st equation}}{\boxed{y+6}+2y=12\implies 3y+6=12} \\\\\\ 3y=6\implies y=\cfrac{6}{3}\implies \blacktriangleright y=2 \blacktriangleleft \\\\\\ \stackrel{\textit{since we know that }}{x=y+6\implies }x=(2)+6\implies \blacktriangleright x=8 \blacktriangleleft \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (8,2)~\hfill$

7 0

3 years ago

Read 2 more answers