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3 years ago

Solve the picture attached

Mathematics

2 answers:

poizon [28]3 years ago

8 0

Answer:

(2, 6)

Step-by-step explanation:

The x value is not going to be changed, but the y value is going to multiply by 1.5. Therefore, the y value would be 4*1.5 = 6.

The answer is (2,6).

Lady_Fox [76]3 years ago

4 0

Your answer is (2,6)

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Solve using an equation: What is 150% of 625.

leonid [27]

Answer:

937.5

Step-by-step explanation:

150% is equal to 150/100. Therefore, we have to multiply 625 by 150/100. 150/100 is the same as 1.5. Therefore, we have to multiply 625 by 1.5. 625x1.5=937.5.

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2 years ago

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a first number plus twice a second number is 4. twice the first number plus the second number totals 26. find the numbers

blagie [28]

Answer:

a = 16

b = -6

Step-by-step explanation:

a + 2b = 4

2a + b = 26

2a + 4b = 8

2a + b = 26

3b = -18

b = -6

a + 2(-6) = 4

a - 12 = 4

a = 16

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2 years ago

Determine which postulate or theorem can be used to prove that LMN NOL

balu736 [363]

Sss similarity theorem

3 0

3 years ago

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Which function has a vertex on the y-axis? f(x) = (x – 2)2 f(x) = x(x + 2) f(x) = (x – 2)(x + 2) f(x) = (x + 1)(x – 2)

strojnjashka [21]

we know that

If the vertex is on the y-axis, then the x-coordinate of the vertex is equal to zero

we are going to verify the vertex of each one of the functions to determine the solution

Remember that

The equation in vertex form of a vertical parabola is equal to

$y=a(x-h)^{2} +k$

where

(h,k) is the vertex

if $a>0$ -------> the parabola open upward (vertex is a minimum)

if $a$ -------> the parabola open downward (vertex is a maximun)

case A) $f(x)=(x-2)^{2}$

This is a vertical parabola open upward

the vertex is the point $(2,0)$

therefore

The function $f(x)=(x-2)^{2}$ does not have a vertex on the y-axis

case B) $f(x)=x(x+2)$

$f(x)=x(x+2)=x^{2}+2x$

convert to vertex form

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+1=x^{2}+2x+1$

Rewrite as perfect squares

$f(x)+1=(x+1)^{2}$

$f(x)=(x+1)^{2}-1$

the vertex is the point $(-1,-1)$

therefore

The function $f(x)=x(x+2)$ does not have a vertex on the y-axis

case C) $f(x)=(x-2)(x+2)$

$f(x)=(x-2)(x+2)=x^{2}-2^{2}$

$f(x)=x^{2}-4$

the vertex is the point $(0,-4)$

The x-coordinate of the vertex is equal to zero

therefore

The function $f(x)=(x-2)(x+2)$ has a vertex on the y-axis

case D) $f(x)=(x+1)(x-2)$

$f(x)=(x+1)(x-2)\\ \\f(x)= x^{2}-2x+x-2 \\ \\f(x)= x^{2} -x-2$

convert to vertex form

Group terms that contain the same variable, and move the constant to the opposite side of the equation

$f(x)+2= x^{2} -x$

Complete the square. Remember to balance the equation by adding the same constants to each side.

$f(x)+2+0.25= x^{2} -x+0.25$

$f(x)+2.25= x^{2} -x+0.25$

Rewrite as perfect squares

$f(x)+2.25= (x-0.50)^{2}$

$f(x)=(x-0.50)^{2}-2.25$

the vertex is the point $(0.5,-2.25)$

therefore

The function $f(x)=(x+1)(x-2)$ does not have a vertex on the y-axis

the answer is

$f(x)=(x-2)(x+2)$

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3 years ago

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Square ABCD and isosceles triangle BUC are drawn to create trapezoid AUCD. Square A B C D and triangle B U C are attached at sid

Art [367]

Answer:

d) 135º

Step-by-step explanation:

Note that the angle DCU is the sum of the angles DCB and BCU. The angle DCB is 90º because A B C D is a square, then all its angles are equal to 90º.

After attaching B U C to A B C D, we obtain a trapezoid A U C D. Since A U C D has at least one pair of parallel sides, then AU should be parallel to CD, thus the angle CBU must be 90º.

B U C is isoceles, so we conclude that other two angles must have the same size, and due to the sum of the angles of a triangle being 180º, then both BUC and BCU are equal to 45º

As a result, the angle DCU is equal to 90º+45º = 135º. Option d is the correct one.

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3 years ago

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