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Gala2k [10]
3 years ago
5

What is the height of parallelogram with base 6.75 and area of 218.7

Mathematics
2 answers:
lara [203]3 years ago
7 0
Area of parallelogram= base x height
218.7= 6.75 x height
height= 218.7÷6.75
height= 32.4
netineya [11]3 years ago
7 0
The height of a parallelogram with a base of 6.75 and an area of 218.7 the height would be: 32.4.

Hope I helped!

- Amber
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Sin =9/15 . Find tan 0
ELEN [110]

Answer:

c. 9/12

Step by step explanation.

Given: sin = 9/15

and,

sin =  \frac{opp.}{hyp.}

tan =  \frac{opp.}{adj.}

thus, find adj. side

by using Pythagoras theorem

{c }^{2} =   {a}^{2}  +  {b}^{2}

then,

let adj. side be a^

{a}^{2}  =  {c}^{2}  -  {b }^{2} \\  {a}^{2}   =  {15 }^{2}  -  {9}^{2}  \\  {a}^{2}  = 144 \\ a = 12

thus,

adj. side is 12

then,

tan of the angle = 9/12

3 0
3 years ago
the sides of a triangle, in centimetres,are x,2x-3 and 2x+5. The perimeter of the triangle is 57 cm.a)write down an equation for
soldier1979 [14.2K]

Answer:

11 cm, 19 cm, 27 cm

Step-by-step explanation:

Triangles have 3 sides and perimeter is the sum of all side lengths. So the equation is x + (2x-3) + (2x+5) = 57.

x + (2x-3) + (2x+5) = 57

x + 2x - 3 + 2x + 5 = 57

x + 2x + 2x - 3 + 5 = 57

5x + 2 = 57

5x = 57 -2

5x = 55

x = 11

Now that we know the value of x, we can figure out the side lengths. The first side (x) is 11 cm long, the second side (2x - 3) is 19 cm, and the third side (2x + 5) is 27 cm long.

6 0
3 years ago
Read 2 more answers
option 1 drop down are: even-odd identity, quotient identity, Pythagorean identity, double-number identity.option 2 drop down ar
motikmotik

Answer:

The equation is given below as

\frac{\cos2x}{\cos x}=\cos x-\sin x\tan x

Step 1:

We will work on the left-hand side, we will have

\begin{gathered} \cos x-\sin x\tan x \\ \text{recall that,} \\ Quoitent\text{ identity is} \\ \tan x=\frac{\sin x}{\cos x} \end{gathered}

By substituting the identity above, we will have

\begin{gathered} \cos x-\sin x\tan x=\cos x-\frac{\sin x.\sin x}{\cos x}=\cos x-\frac{\sin^2x}{\cos x} \\  \end{gathered}

Here, we will make use of the quotient identity

Step 2:

By writings an expression, we will have

\begin{gathered} \cos x-\sin x\tan x=\cos x-\frac{\sin x.\sin x}{\cos x} \\ \cos x-\sin x\tan x=\frac{\cos^2x-\sin^2x}{\cos x} \end{gathered}

Here, we will use the definition of subtraction

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Step 3:

We will apply the double number identity given below

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By applying this, we will have

\frac{\cos^2x-\sin^2x}{\cos x}=\frac{\cos2x}{\cos x}

Here, we will use the double number identity

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5 0
1 year ago
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You haven't shared the given line, so all I can do here is to invent a line and then show you how to write the equation of a new line which is parallel to mine and which has an x-intercept of 4.


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The new line MUST have the same slope: m = 2/3.


Then y = mx + b becomes y = (2/3)x + b. Find the x-intercept by setting y = 0 and solving for x: (2/3)x = 0 - b. Now replace x with 4 and find b:


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y = (2/3)x - 8/3.

3 0
3 years ago
Read 2 more answers
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