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3 years ago

Can u help me with a math question plz?

Mathematics

1 answer:

Furkat [3]3 years ago

6 0

Given: 48 students took the exam.

A total of 18 + 10 = 28 students scored 80% or above.

To find the percentage, take the part (28 students) divided by the whole (48 students) then multiply the quotient by 100.

(28 students / 48 students) x 100 = 58.33%

58.33% of the students who took the exam scored an 80% or higher

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The sets A,B and C in a universal set consisting of the first 10 lower case letters of the alphabet are: A={a,b,c}, B={b,c,a,e},

Papessa [141]

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Which item fits in the left side of the table?Criminal versus Civil Cases Criminal cases Civil cases ? . • Accuser is called the plaintiff • Lower burden of proof Defendant may be found liable • Usually heard by a judge only • Accused not guaranteed an attorney

A. No burden of proof

B. Accuser is called the defendant

C. Defendant may be found guilty

D. Usually heard by a jury only

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3 years ago

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Round decimals <br> Write the place value of the underline digit

atroni [7]

So all you have to do is look at were the number is is it is in the first spot it is in the tenths place value, second spot hundredth and so

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3 years ago

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.

4 0

3 years ago

Will give brainliest

Dovator [93]

Answer:

GURL MY MATH TEACHER GAVE ME THE SAME SHEET A MONTH AGOO

YASS

1.) x is less than or equal to 3

2.) x greater than or equal to 2

3.) x < -1

4.) x greater than or equal to 3

5.) x>2

6.) x greater than or equal to 1

5 0

3 years ago