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2 years ago

The intersection of a triangle and a line could be a segment True or False

Mathematics

1 answer:

Arlecino [84]2 years ago

6 0

false

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Please help! If possible please show your work :D Thank you! Solve.

nataly862011 [7]

Equation at the end of step1: (7x + 18) (4) - 4x = 0 2 STEP2:Equation at the end of step 2 2 • (7x + 18) - 4x = 0 STEP3:STEP4:Pulling out like terms

4.1 Pull out like factors :

10x + 36 = 2 • (5x + 18)

Equation at the end of step4: 2 • (5x + 18) = 0 STEP5:Equations which are never true:

5.1 Solve : 2 = 0

This equation has no solution.

A a non-zero constant never equals zero.

Solving a Single Variable Equation:

5.2 Solve : 5x+18 = 0

Subtract 18 from both sides of the equation :

5x = -18

Divide both sides of the equation by 5:

x = -18/5 = -3.600

6 0

2 years ago

Solve the following equations;- x - 9 /√x + 3 =1

fgiga [73]

Answer:

13 , 6.

Step-by-step explanation:

A equation is given to us and we need to find out the value of x . The given equation to us is ,

$\sf\implies \dfrac{ x -9}{\sqrt{x+3}}= 1$

Cross multiply ,

$\sf\implies x - 9 =\sqrt{ x +3}$

Squaring both sides ,

$\sf\implies (x - 9)^2 = x + 3$

Simplify the whole square ,

$\sf\implies x^2+9^2-2.9.x = x +3 \\$

$\sf\implies x^2+81 - 18x = x -3$

Add 3 and subtract x on both sides ,

$\sf\implies x^2 -18x-x +81-3=0$

Simplify ,

$\sf\implies x^2 -19x +78=0$

Split the middle term of the quadratic equation,

$\sf\implies x^2-13x-6x+78=0$

Take out common ,

$\sf\implies x( x -13)-6(x-13)=0$

Take out ( x -13 ) as common ,

$\sf\implies ( x -13)(x-6)=0$

Equate both factors to 0 ,

$\sf\implies \boxed{\pink{\sf x = 13,6}}$

Hence the value of x is 13 or 6 .

7 0

2 years ago

What is the solution set for z+10=z-2

photoshop1234 [79]

Answer: No Solution.

There are no values of z that make the equation true.

Step-by-step explanation:

5 0

2 years ago

Explain in a minimum of 2 sentences how to graph the equation of the absolute value function given a vertex of (-1,3) and a valu

VladimirAG [237]

This would be y = |x+1| + 3

7 0

2 years ago

Write the quadratic equation whose roots are 5 and 1, and whose leading coefficient is 5

zepelin [54]

Answer: the equation is

5x^2 -30x - 25

Step-by-step explanation:

A quadratic equation is one in which the highest power of the unknown is 2.

The general form of a quadratic equation is expressed as

ax^2 + bx + c

Where c is a constant and a is the leading coefficient

Assuming we want to write the quadratic equation in x, from the information given, the given roots are 5 and 1 and the leading coefficient is 5. We will just multiply the expression by the leading coefficient.

Therefore, the linear factors of the quadratic will be (x-5) and (x-1)

With the leading coefficient as 5, the equation becomes

5(x-5)(x-1)

= 5(x^2 - x - 5x + 5)

= 5(x^2 - 6x + 5)

= 5x^2 -30x - 25

3 0

2 years ago