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3 years ago

The side lengths of a triangle are increased by 50% is the area increased by 50% as well

1 answer:

3 0

No. The area of a triangle is Base*Height/2 so if both base and height increased by 50%, the area would equal (Base*1.5)*(Height*1.5)/2, which is 2.25 bigger than the original triangle, not 1.5 times bigger(50% more)

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Ana earns 75 points and loses 30 points write and solve an expression to find anas final score

steposvetlana [31]

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3 years ago

Write the equation of the line fully simplified slope-intercept form.

Pepsi [2]

Answer:

1234567890

Step-by-step explanation:

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3 years ago

Find the values of A B C AND D of 4x^2 (2x^3+5x)= Ax^B +Cx^D

iren2701 [21]

The values are $A=8, B=5, C= 20$ and $D=3$

Explanation:

The expression is $4x^{2} (2x^{3} +5x)=Ax^{B} +Cx^{D}$

Simplifying, we get,

$8x^{5} +20x^3=Ax^{B} +Cx^{D}$

Since, both sides of the expression are equal, we can equate the corresponding values of A, B, C and D.

Thus, we get,

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Also, equating, $20 x^{3}=C x^{D}$ , we get,

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3 0

3 years ago

Please help me solve this for 15 points! -x2 + 4x - 54 = 0

Mazyrski [523]

Answer:

x1 =2-5i*sqrt(2)

x2 =2+5i*sqrt(2)

Step-by-step explanation:

-x^2 +4x-54=0 (quadratic equation)

a=-1, b=4, c=-54

x1=(-b+sqrt(b^2-4ac))/2a

x1=(-4+sqrt(4^2 - 4*(-1)(-54))/2*(-1)

x1=(-4+sqrt(16-216))/(-2)

x1 =(-4+sqrt(-200))/(-2)

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x1 =(-4+sqrt(100*2*i^2))/(-2)

x1 =(-4+10i*sqrt(2))/(-2)

x1 =2-5i*sqrt(2)

x2 =(-b-sqrt(b^2-4ac))/2a

x2 =(-4-10i*sqrt(2))/(-2)

x2 =2+5i*sqrt(2)

7 0

3 years ago

Read 2 more answers

Solve the equation. 12a=11

USPshnik [31]

A= 11/12 i hope this helped !!!

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2 years ago