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3 years ago

I need help! can anybody help?

Mathematics

1 answer:

Alexandra [31]3 years ago

5 0

Answer:

i, m, and then k (use my method to do the other one)

Step-by-step explanation:

The longer than angle the longer the side tht is opposite to it... so for example, each angle and its opposite correspond to each other.. so K it is side K and so on...

first u should find the missing angles, for A, we know the angles 118 and 28 and a triangle is a total of 180 degrees, so add 118 and 28, which will give u 146, now subtract tht from 180 and u get 34, so the missing angle M is 34, now we can see which angles correspond to which side:

K = k

M = m

L = l

so the shortest will be l, then it will be m, and then it will be k

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Tess made a mosaic in art class with different-shaped tiles. She started by putting 2 rows of t triangle tiles at the top of the

borishaifa [10]

Answer:

you mean Tessa right

Step-by-step explanation:

get it, because my name is tessa? lol

7 0

4 years ago

Help!!! Please help

Flauer [41]

It's blurry on my screen please take it closer

8 0

3 years ago

Read 2 more answers

Help!! Answer !!! about to run out of time in test!!

Gemiola [76]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

Here's the solution ~

Let's find the measure of hypotenuse first, by using Pythagoras theorem ;

$\qquad \sf \dashrightarrow \: h {}^{2} = {8}^{2} + {6}^{2}$

$\qquad \sf \dashrightarrow \: h {}^{2} = {36}^{} + {64}^{}$

$\qquad \sf \dashrightarrow \: h {}^{2} = 100$

$\qquad \sf \dashrightarrow \: h {}^{} = \sqrt{100}$

$\qquad \sf \dashrightarrow \: h {}^{} = {10}$

Now, let's find the asked values ~

$\qquad \sf \dashrightarrow \: \sin(x) = \dfrac{opposite \: side}{hypotenuse}$

$\qquad \sf \dashrightarrow \: \sin(x) = \dfrac{6}{10}$

$\qquad \sf \dashrightarrow \: \sin(x) = \dfrac{3}{5} \: or \: 0.6 \: units$

For Cos y :

$\qquad \sf \dashrightarrow \: \cos(y) = \dfrac{adjcant \: side}{hypotenuse}$

$\qquad \sf \dashrightarrow \: \cos(y) = \dfrac{6}{10}$

$\qquad \sf \dashrightarrow \: \cos(y) = \dfrac{3}{5} \: or \: 0.6 \: units$

As we can see that both sin x and Cos y have equal values, therefore The required relationships is equality.

I.e Sin x = Cos y

Hope it helps ~

3 0

3 years ago

Parallel / Perpendicular Practice

deff fn [24]

The slope and intercept form is the form of the straight line equation that includes the value of the slope of the line

Neither
║
Neither
⊥
║
Neither
Neither
Neither

Reason:

The slope and intercept form is the form y = m·x + c

Where;

m = The slope

Two equations are parallel if their slopes are equal

Two equations are perpendicular if the relationship between their slopes, m₁, and m₂ are; $m_1 = -\dfrac{1}{m_2}$

1. The given equations are in the slope and intercept form

$\ y = 3 \cdot x + 1$

The slope, m₁ = 3

$y = \dfrac{1}{3} \cdot x + 1$

The slope, m₂ = $\dfrac{1}{3}$

Therefore, the equations are neither parallel or perpendicular

Neither

2. y = 5·x - 3

10·x - 2·y = 7

The second equation can be rewritten in the slope and intercept form as follows;

$y = 5 \cdot x -\dfrac{7}{2}$

Therefore, the two equations are parallel

3. The given equations are;

-2·x - 4·y = -8

-2·x + 4·y = -8

The given equations in slope and intercept form are;

$y = 2 -\dfrac{1}{2} \cdot x$

Slope, m₁ = $-\dfrac{1}{2}$

$y = \dfrac{1}{2} \cdot x - 2$

Slope, m₂ = $\dfrac{1}{2}$

The slopes

Therefore, m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

The lines are Neither parallel nor perpendicular

Neither

4. The given equations are;

2·y - x = 2

$y = \dfrac{1}{2} \cdot x +1$

m₁ = $\dfrac{1}{2}$

y = -2·x + 4

m₂ = -2

Therefore;

$m_1 \neq -\dfrac{1}{m_2}$

Therefore, the lines are perpendicular

5. The given equations are;

4·y = 3·x + 12

-3·x + 4·y = 2

Which gives;

First equation, $y = \dfrac{3}{4} \cdot x + 3$

Second equation, $y = \dfrac{3}{4} \cdot x + \dfrac{1}{2}$

Therefore, m₁ = m₂, the lines are parallel

6. The given equations are;

8·x - 4·y = 16

Which gives; y = 2·x - 4

5·y - 10 = 3, therefore, y = $\dfrac{13}{5}$

Therefore, the two equations are neither parallel nor perpendicular

Neither

7. The equations are;

2·x + 6·y = -3

Which gives $y = -\dfrac{1}{3} \cdot x - \dfrac{1}{2}$

12·y = 4·x + 20

Which gives

$y = \dfrac{1}{3} \cdot x + \dfrac{5}{3}$

m₁ ≠ m₂

$m_1 \neq -\dfrac{1}{m_2}$

Neither

8. 2·x - 5·y = -3

Which gives; $y = \dfrac{2}{5} \cdot x +\dfrac{3}{5}$

5·x + 27 = 6

$x = -\dfrac{21}{5}$

Therefore, the slopes are not equal, or perpendicular, the correct option is Neither

Learn more here:

brainly.com/question/16732089

6 0

3 years ago

4x-3=5x-21 what does x=

torisob [31]

U subtract 4x by 5x and then you add 3 by -21 and the answer is x=18

3 0

4 years ago

I need help! can anybody help?​

I need help! can anybody help?