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3 years ago

Need help with third question!!!!!!!!!!

Mathematics

2 answers:

Leokris [45]3 years ago

5 0

Answer:

Angle 2 = Angle 4 (alternate angles)
Angle 1 = Angle 5 (corresponding angles)

Angle 1 + Angle 2 + Angle 3 = 180 (angles on a line)

Now,

Angle 2 = Angle 4 (alternate angles)

and Angle 1 = Angle 5 (corresponding angles)

So,

Angle 1 + Angle 2 + Angle 3 = Angle 5 + Angle 4 + Angle 3 = 180

Hence proved

Step-by-step explanation:

Hope this helps

s344n2d4d5 [400]3 years ago

5 0

Yea what the person above said.

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The length of a rectangle is 5 m less than twice the width, and the area of the rectangle is 52 m^2. Find the dimensions of the

lys-0071 [83]

length is 8m

Width is 6.5m

Explanation:

Let, width = b

So, length = 2b-5

area = 52 $m^{2}$

We know,

area of rectangle = length X width

So, $52m^{2}$ = $(2b-5) X b$

On solving the above equation, we get

Width, b = 6.5m

length, l = 8m

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Alenkasestr [34]

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To two decimal places, find the value of k that will make the function f(x) continuous everywhere. f of x equals the quantity 3x

Fiesta28 [93]

Given function is

$f(x)=\left\{\begin{matrix}3x+k & x\leq -4 \\ kx^2-5 & x> -4\end{matrix}\right.$

now we need to find the value of k such that function f(x) continuous everywhere.

We know that any function f(x) is continuous at point x=a if left hand limit and right hand limits at the point x=a are equal.

So we just need to find both left and right hand limits then set equal to each other to find the value of k

To find the left hand limit (LHD) we plug x=-4 into 3x+k

so LHD= 3(-4)+k

To find the Right hand limit (RHD) we plug x=-4 into

$kx^2-5$

so RHD= $k(-4)^2-5$

Now set both equal

$k(-4)^2-5=3(-4)+k$

$16k-5=-12+k$

$16k-k=-12+5$

$15k=-7$

$k=-\frac{7}{15}$

k=-0.47

Hence final answer is -0.47.

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3 years ago