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2 years ago

15

Please help me. Noura is redecorating her house. She needs to work out the area of the wall around her triangular window in orde

r
to find out how much paint she
needs.

a) Calculate the area of the wall
she needs to paint, to a suitable level
of accuracy, explain your answer.
b) Paint is sold in 20 litres containers,
costing $21.99 per container. Use
your answer from part (a) to
calculate and explain how much it
will cost to paint the wall. Assume 1
litre of paint covers 1m2 *

1 answer:

Vinvika [58]2 years ago

8 0

Answer:

a. 277.3 m²

b. $307.86

Step-by-step explanation:

a. Area of the wall to be painted = area of rectangle - area of triangle

= L × W + ½×a×b×sin θ

L = 17 m

W = 14 m

a = 8.7 m

b = 9.5 m

θ = 72°

Plug in the values into the equation

Area of the wall = (17×14) + (½*8.7×9.5×sin 72)

Area of the wall = 238 + 39.3024105

Area of the wall to paint ≈ 277.3 m²

b. 20 liters of paint of 1 container cost $21.99

If 1 liter of paint covers 1m², therefore,

277.3 m² will need = 277.3 × 1 = 277.3 liters of paint.

20 liters = 1 container of paint

277.3 liters = 277.3/20 = 13.865 ≈ 14 containers

1 container = $21.99

14 containers = 21.99 × 14 = $307.86

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Solve the equation for<br> x 3/4(x+21) = 10.5 <br> x =

MrRa [10]

<h3>✍️ Answer:</h3><h3> $x=-\frac{294}{13}$ </h3>

✽✽✽✽

$\frac{x\cdot \:3}{4\left(x+21\right)}=10.5$

$\mathrm{Multiply\:both\:sides\:by\:}4\left(x+21\right)$

$\frac{x\cdot \:3}{4\left(x+21\right)}\cdot \:4\left(x+21\right)=10.5\cdot \:4\left(x+21\right)$

$3x=42\left(x+21\right)$

$3x=42x+882$

$\mathrm{Subtract\:}42x\mathrm{\:from\:both\:sides}$

$3x-42x=42x+882-42x$

$-39x=882$

$\mathrm{Divide\:both\:sides\:by\:}-39$

$\frac{-39x}{-39}=\frac{882}{-39}$

$x=-\frac{294}{13}$

❅❅❅❅❅❅❅❅❅❅

<h3>hope it helps...</h3><h3>have a great day!!</h3>

4 0

2 years ago

In this diagram, ABAC – AEDF. If the<br> area of ABAC = 6 in?, what is the<br> area of AEDF?

shepuryov [24]

Answer:

2.7 in²

Step-by-step explanation:

similar triangles have the same angles, and all side lengths (or other distances) of one triangle have the same scaling factor to the side lengths of the other triangle.

so, we know the relation between the 2 baselines is 2/3, as this is the factor to turn the baseline of the large triangle into the baseline of the smaller triangle.

in other words

EF = BC × 2/3

2 = 3 × 2/3

correct

how do we calculate the area of a triangle ?

Area = baseline × height / 2

from BAC we know

Area = 6

baseline = 3

height = ?

6 = 3 × height / 2

12 = 3 × height

height = 4

aha !

now, EDF has a height too that we need to calculate is Area. and this height has the same scaling factor compared to the larger triangle as the side lengths : 2/3

so, for EDF we know

Area = ?

baseline = 2

height = 4 × 2/3 = 8/3

therefore, the area is

Area = (2 × 8/3) / 2 = (16/3) / 2 = 8/3 = 2.66666... ≈ 2.7

the shirt answer would be :

we know from the 2 baselines that the scaling factor for each distance is 2/3.

for the area we need to multiply 2 distances, so that means we have to multiply both by 2/3. and so on the formula for the area we have to use 2/3 × 2/3.

2/3 × 2/3 = 4/9

=>

Area small = Area large × 4/9 = 6 × 4/9 = 24/9 = 8/3 ≈ 2.7

3 0

2 years ago

What is the value of x in: 4x+3y=12 and 2x-3y=-30 *

NNADVOKAT [17]

Answer:

The value would be

Step-by-step explanation:

What is the value of x in: 4x+3y=12 and 2x-3y=-30 *

4x + 3y

The answer is x = 2 - 3y = 4

3y = 30

So 2x + 3y = 12 is

X = 3 - 3y/4

2x - 3y = 30

X = 15 + 3y/2

Hope it help : ) Please Rate it for me!

Have a Great Day and Winter Break!

6 0

2 years ago

Leya [2.2K]

Answer:

w=-7/2z

Step-by-step explanation:

to solve for w you firstly have to group the like terms

5w-3w=2z-9z

2w=-7z

then you have to divide both sides by 2 inorder to remain with w on one side.

2w/2=-7/2

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I hope this helps

3 0

2 years ago

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Which of the following is not equal to sin⁡(-230°)?

xenn [34]

From trigonometry we know that:

if $sin(\theta)=sin(\alpha)$

then, $\theta=n\pi+(-1)^n\alpha$ (where $n$ is an integer)

This can be rewritten in degrees as:

$\theta=n(180^{\circ})+(-1)^n\alpha$ .............(Equation 1)

Now, in our case, $\alpha=-230^{\circ}$

Therefore, (Equation 1) can be written as:

$\theta=n(180^{\circ})+(-1)^n(-230^{\circ})$ ..........(Equation 2)

Now, to find the correct options all that we have to do is replace n by relevant integers and find the values of $\theta$ that match.

For n=2, (Equation 2) gives us: $\theta=2\times 180^{\circ}+(-1)^2(-230^{\circ})=360^{\circ}-230^{\circ}=130^{\circ}$ .

Thus, $sin(230^{\circ})=sin(130^{\circ})$

Now, we know that: $-sin(-50^{\circ})=sin(50^{\circ})$

Let n=-1, then:

$\theta=(-1)\times 180^{\circ}+(-1)^{-1}(-230^{\circ})=-180^{\circ}+230^{\circ}=50^{\circ}$

Thus, $sin(-230^{\circ})=-sin(-50^{\circ})$

Likewise, $sin(-230^{\circ})=sin(50^{\circ})$

Only the last option $sin(-50^{\circ})$ will never match $sin(-230^{\circ})$ because no integral value of $n$ will ever give $\theta=-50^{\circ}$

Thus the last option is the correct option.

3 0

2 years ago

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