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Rzqust [24]
3 years ago
14

What values of b will cause 4x^2+bx+25=0 to have one real solution?

Mathematics
1 answer:
Darina [25.2K]3 years ago
5 0

Answer:

b=20 or b=-20

Step-by-step explanation:

Keep in mind the meanings of the values of the discriminant:

If b^2-4ac=0, then the quadratic will have only 1 solution (double root)

If b^2-4ac<0, then the quadratic will have no real solutions

If b^2-4ac>0, then the quadratic will have 2 unique solutions

In this case, to get one real solution, the discriminant must be set up as b^2-4ac=0. Setting up the equation we get:

b^2-4(4)(25)=0

b^2-400=0

b^2=400

b=20 or b=-20

So the values of b would have to be either 20 or -20

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The diagram below represent the rate of making copies on a copy machine. Which table represent this rate?
Nuetrik [128]

Answer:

its B

Step-by-step explanation:

6 0
2 years ago
Please help me with this homework question. No silly answers please and no bots! I will mark brainliest to whoever gets all them
MatroZZZ [7]

Answer:

The formula for the area of the triangle is:

\displaystyle A=24x^2

When <em>x</em> = 3, the area is 216 square centimeters.

When <em>x </em>= 2.4, the area is 138.24 square centimeters.

And when <em>x</em> = 5, the hypotenuse of the triangle is 50 centimeters.

Step-by-step explanation:

We are given a right triangle with a height of 8<em>x</em> and a base of 6<em>x</em>.

The area of a triangle is given by the formula:

\displaystyle A=\frac{1}{2}bh

So, by substitution:

\displaystyle A=\frac{1}{2}(8x)(6x)=24x^2

When <em>x </em>= 3, the area will be:

\displaystyle A(3)=24(3)^2=216\text{ cm}^2

And when <em>x</em> = 2.4, the area will be:

A(2.4)=24(2.4)^2=138.24\text{ cm}^2

When <em>x </em>= 5, the height will be 8(5) = 40 cm and the base will be 6(5) = 30 cm.

By the Pythagorean Theorem:

a^2+b^2=c^2

So:

(40)^2+(30)^2=c^2

The hypotenuse will be:

c=\sqrt{1600+900}=\sqrt{2500}=50\text{ cm}

3 0
3 years ago
What are the coordinates of the orthocenter of △JKL with vertices at J(−4, −1) , K(−4, 8) , and L(2, 8) ?
Sati [7]

<u>Answer-</u>

<em>The coordinates of the orthocenter of △JKL is (-4, 8)</em>

<u>Solution-</u>

The orthocenter is the point where all three altitudes of the triangle intersect. An altitude is a line which passes through a vertex of the triangle and is perpendicular to the opposite side.

For a right angle triangle, the vertex at the right angle is the orthocentre of the triangle.

Here we are given the three vertices of the triangle are  J(-4,-1), K(-4,8) and L(2,8)

If the triangle JKL satisfies Pythagoras Theorem, then triangle JKL will be a right angle triangle.

Applying distance formula we get,

JK^2= (-4+4)^2+ (8+1)^2=0+81=81\\\\KL^2= (-4-2)^2+ (8-8)^2=36+0=36\\\\JL^2= (-4-2)^2+(8+1)^2=36+81=117

As,

\Rightarrow 117=81+36

\Rightarrow JL^2=JK^2+KL^2

\Rightarrow \text{JKL is a right angle triangle}

\Rightarrow \angle K=90^{\circ}

Therefore, the vertex at K (-4, 8) is the orthocentre.

3 0
3 years ago
What’s is the correct answer???
dangina [55]

The correct answer would be c. Hope this helps!

5 0
3 years ago
Please help i really need it !!!!
Vika [28.1K]
C, i think you forgot an x
(x+11)+(x-7)
x+11+x-7
x+x+4
2x+4
4 0
3 years ago
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