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Paul [167]
3 years ago
10

I need help please thanks!

Mathematics
1 answer:
Natasha_Volkova [10]3 years ago
7 0

Answer:

f=39

Step-by-step explanation:

hi! first, we can multiply by 12 on both sides to eliminate the 12 in the denominator.

f+3= (7/2)*12

f+3=84/2

f+3=42

now, subtract 3 on both sides.

f=39

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¯¯¯¯¯¯ J K is a tangent to circle C . If m ∠ K J L = 27 °, What is m ˆ K L ?
iragen [17]

Answer: I think 13

Step-by-step explanation:

5 0
3 years ago
The line passing through (-2,5) and (2,p) has a gradient of -1/2. <br><br> Fnd the value of p
makkiz [27]

Given:

The line passing through (-2,5) and (2,p) has a gradient of -\dfrac{1}{2}.

To find:

The value of p.

Solution:

If a line passes through two points, then the slope of the line is:

m=\dfrac{y_2-y_1}{x_2-x_1}

The line passing through (-2,5) and (2,p). So, the slope of the line is:

m=\dfrac{p-5}{2-(-2)}

m=\dfrac{p-5}{2+2}

m=\dfrac{p-5}{4}

It is given that the gradient or slope of the line is -\dfrac{1}{2}.

\dfrac{p-5}{4}=-\dfrac{1}{2}

On cross multiplication, we get

2(p-5)=-4(1)

2p-10=-4

2p=-4+10

2p=6

Divide both sides by 2.

p=3

Therefore, the value of p is 3.

4 0
3 years ago
Which algebraic expression is equivalent to the expression below 6(2x+5)+4x
Veronika [31]
12x+30+4
12x+34

This is 12x + 34 because when we combine all like terms, we get it.
6 0
3 years ago
Instructions: Solve the following literal equation.<br> Solve P-2w/2= l for P.
mrs_skeptik [129]

Answer:

P=l+w

Step-by-step explanation:

Solve for P by simplifying both sides of the equation, then isolating the variable.

6 0
3 years ago
The figures below are made out of circles, semicircles, quarter circles, and a square. Find the area and the perimeter of each f
WITCHER [35]

Answer:

Part 1) A=36(\pi-2)\ cm^2

Part 2) P=6(\pi+2\sqrt{2})\ cm

Step-by-step explanation:

<u><em>The picture of the question in the attached figure</em></u>

Part 1) Find the area

we know that

The area of the shape is equal to the area of a quarter of circle minus the area of an isosceles right triangle

so

A=\frac{1}{4}\pi r^{2}-\frac{1}{2}(b)(h)

we have that the base and the height of triangle is equal to the radius of the circle

r=12\ cm\\b=12\ cm\\h=12\ cm

substitute

A=\frac{1}{4}\pi (12)^{2}-\frac{1}{2}(12)(12)\\A=(36\pi-72)\ cm^2

simplify

Factor 36

A=36(\pi-2)\ cm^2

Part 2) Find the perimeter

The perimeter of the figure is equal to the circumference of a quarter of circle plus the hypotenuse of the right triangle

The circumference of a quarter of circle is equal to

C=\frac{1}{4}(2\pi r)=\frac{1}{2}\pi r

substitute the given values

C=\frac{1}{2}\pi (12)\\C=6\pi\ cm

Applying the Pythagorean Theorem

The hypotenuse of right triangle is equal to

AC=\sqrt{12^2+12^2}\\AC=\sqrt{288}\ cm

simplify

AC=12\sqrt{2}\ cm

Find the perimeter

P=(6\pi+12\sqrt{2})\ cm

simplify

Factor 6

P=6(\pi+2\sqrt{2})\ cm

4 0
3 years ago
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