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3 years ago

Need help , don’t understand !!

Mathematics

1 answer:

mixer [17]3 years ago

5 0

Quadratic Formula is (-b ±√(b²-4ac))/2a

With that being said:

y=2x²-5x-1 is in the correct form of ax²+bx+c

Now just plug in:

5±√(25-4(2 x-1))

5± √25+8

(5±√33)/2(2)

Answer is D

100% Sure

Brainliest?

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The high school has 26 teachers and 546 students. What is the ratio of teachers to students?

skelet666 [1.2K]

Answer:

1:21

Step-by-step explanation:

Easiest trick way to do this is do 26/546 on your calculator (if you got a scientific one) this would simplify this to its lowest fraction which can be converted into a ratio in this case the ratio of teachers to students is 1:21

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3 years ago

HELP!!! giving brainliest

Stolb23 [73]

4cm = 5km

5/4 = 1.25

1cm = 1.25km

5 × 1.25 = 6.25

6.25km

Hope this helps! :)

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4 years ago

Round 0.978 to the nearest tenth

Setler79 [48]

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3 years ago

The Farrells save $150 each month for their next summer vacation.Write an equation that they can use to find ,s, their savings a

Andrei [34K]

Answer:

s=150m

Step-by-step explanation:

Using slope intercept form, you can come p with the equation, s=150m where 150 is the savings per month and you were not given a starting point which would be y.

y=mx+b

s=150m

7 0

3 years ago

Read 2 more answers

Solve the equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are locat

eimsori [14]

Answer:

There are NO real roots for this equation. The only roots have imaginary parts and therefore cannot be represented on the real x-axis.

Step-by-step explanation:

We notice that the expression on the left of the equation is a quadratic with leading term $2x^2$ , which means that its graph is that of a parabola with branches going up.

Therefore, there can be three different situations:

1) if its vertex is ON the x axis, there would be one unique real solution (root) to the equation.

2) if its vertex is below the x-axis, the parabola's branches are forced to cross it at two locations, giving then two real solutions (roots) to the equation.

3) if its vertex is above the x-axis, it will have NO real solutions (roots) but only non-real ones.

So we proceed to examine the vertex's location, which is also a great way to decide on which set of points to use in order to plot its graph efficiently.

We recall that the x-position of the vertex for a quadratic function of the form $f(x)=ax^2+bx+c$ is given by the expression:

$x_v=\frac{-b}{2a}$

Since in our case $a=2$ and $b=-3$ , we get that the x-position of the vertex is:

$x_v=\frac{-b}{2a}\\x_v=\frac{-(-3)}{2(2)}\\x_v=\frac{3}{4}$

Now we can find the y-value of the vertex by evaluating this quadratic expression for x = 3/4:

$y_v=f(\frac{3}{4})=2( \frac{3}{4})^2-3(\frac{3}{4})+4\\f(\frac{3}{4})=2( \frac{9}{16})-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{9}{4}+4\\f(\frac{3}{4})=\frac{9}{8}-\frac{18}{8}+\frac{32}{8}\\f(\frac{3}{4})=\frac{23}{8}$

This is a positive value for y, therefore we are in the situation where there is NO x-axis crossing of the parabola's graph, and therefore no real roots.

We can though estimate a few more points of the parabola's graph in order to complete the graph as requested in the problem. For such we select a couple of x-values to the right of the vertex, and a couple to the right so we can draw the branches. For example: x = 1, and x = 2 to the right; and x = 0 and x = -1 to the left of the vertex:

$f(-1) = 2(-1)^2-3(-1)+4= 2+3+4=9\\f(0)=2(0)^2-3(0)+4=0+0+4=4\\f(1)=2(1)^2-3(-1)+4=2-3+4=3\\f(2)=2(2)^2-3(2)+4=8-6+4=6$

See the graph produced in the attached image.

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3 years ago