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3 years ago

PLEASE HELP ME ON THIS!!

Mathematics

1 answer:

Ganezh [65]3 years ago

3 0

Answer:

-6

Step-by-step explanation:

This means what value can I plug into -3x-8 so that I get output 10.

g(x)=-3x-8

g(a)=-3a-8

So we are going to solve g(a)=10 for a.

g(a)=10

-3a-8=10

Add 8 on both sides:

-3a =18

Divide both sides by -3:

a =-6

Check it!

g(-6)=-3(-6)-8=18-8=10 and it's good! :)

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ch4aika [34]

Answer:

Step-by-step explanation:

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3 years ago

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If the area around the vegetable garden is of uniform width (labeled with x) and the dimensions of the vegetable garden is 45 fe

Nady [450]

Area of Rectangle = Width × Length

Vegetable Garden
Width = 20
Length = 45
Area of Vegetable Garden = 20×45=900 Sq ft

Area of Entire Garden = Width × Length
Width = 20 + 2x
Length = 45 + 2x
Area = (20 + 2x)(45 + 2x)
(2x + 20)(2x + 45)
4x^2 + 90x + 45x + 900
Area of Entire Garden = 4x^2 + 135x + 900

Area of Flower Garden = Area of Entire Garden - Area of Vegetable Garden

(4x^2 + 135x + 900) - (20 × 45)
4x^2 + 135x + 900 - 900
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3 years ago

Match each set of vertices with the type of triangle they form.

Andrew [12]

Answer: The calculations are done below.

Step-by-step explanation:

(i) Let the vertices be A(2,0), B(3,2) and C(5,1). Then,

$AB=\sqrt{(2-3)^2+(0-2)^2}=\sqrt{5},\\\\BC=\sqrt{(3-5)^2+(2-1)^2}=\sqrt{5},\\\\CA=\sqrt{(5-2)^2+(1-0)^2}=\sqrt{10}.$

Since, AB = BC and AB² + BC² = CA², so triangle ABC here will be an isosceles right-angled triangle.

(ii) Let the vertices be A(4,2), B(6,2) and C(5,3.73). Then,

$AB=\sqrt{(4-6)^2+(2-2)^2}=\sqrt{4}=2,\\\\BC=\sqrt{(6-5)^2+(2-3.73)^2}=\sqrt{14.3729},\\\\CA=\sqrt{(5-4)^2+(3.73-2)^2}=\sqrt{14.3729}.$

Since, BC = CA, so the triangle ABC will be an isosceles triangle.

(iii) Let the vertices be A(-5,2), B(-4,4) and C(-2,2). Then,

$AB=\sqrt{(-5+4)^2+(2-4)^2}=\sqrt{5},\\\\BC=\sqrt{(-4+2)^2+(4-2)^2}=\sqrt{8},\\\\CA=\sqrt{(-2+5)^2+(2-2)^2}=\sqrt{9}.$

Since, AB ≠ BC ≠ CA, so this will be an acute scalene triangle, because all the angles are acute.

(iv) Let the vertices be A(-3,1), B(-3,4) and C(-1,1). Then,

$AB=\sqrt{(-3+3)^2+(1-4)^2}=\sqrt{9}=3,\\\\BC=\sqrt{(-3+1)^2+(4-1)^2}=\sqrt{13},\\\\CA=\sqrt{(-1+3)^2+(1-1)^2}=\sqrt 4.$

Since AB² + CA² = BC², so this will be a right angled triangle.

(v) Let the vertices be A(-4,2), B(-2,4) and C(-1,4). Then,

$AB=\sqrt{(-4+2)^2+(2-4)^2}=\sqrt{8},\\\\BC=\sqrt{(-2+1)^2+(4-4)^2}=\sqrt{1}=1,\\\\CA=\sqrt{(-1+4)^2+(4-2)^2}=\sqrt{13}.$

Since AB ≠ BC ≠ CA, and so this will be an obtuse scalene triangle, because one angle that is opposite to CA will be obtuse.

Thus, the match is done.

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4 years ago

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Please can somebody help me with this question please...!

olga55 [171]

Since you are given that the student registered early, the total number you deal with is all students who registered early.

211 + 329 = 540

number of undergraduates who registered early = 211

Among students who registered early:

p(undergraduate) = 211/540

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4 years ago

How is writing expressions with variables and numbers similar to writing expressions using words?

Vladimir79 [104]

Answer: Similarly, when we describe an expression in words that includes a variable, we're describing an algebraic expression (an expression with a variable). One of the many reasons is that math is more precise and easier to work with than words are.

Step-by-step explanation:

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3 years ago