Find the area of the trapezoid.

The quotient of a number and 5 is less than 4

cupoosta [38]

Answer:

n ÷ 5 < 4

Step-by-step explanation:

Quotient is to divide.

Let n be the unknown number.

Now you have the equation needed.

8 0

3 years ago

Determine the number of real solutions for each of the given equations. Equation Number of Solutions y = -3x2 + x + 12 y = 2x2 -

rosijanka [135]

Answer:

Step-by-step explanation:

Our equations are

$y = -3x^2 + x + 12\\y = 2x^2 - 6x + 5\\y = x^2 + 7x - 11\\y = -x^2 - 8x - 16\\$

Let us understand the term Discriminant of a quadratic equation and its properties

Discriminant is denoted by D and its formula is

$D=b^2-4ac\\$

Where

a= the coefficient of the $x^{2}$

b= the coefficient of $x$

c = constant term

Properties of D: If D

i) D=0 , One real root

ii) D>0 , Two real roots

iii) D<0 , no real root

Hence in the given quadratic equations , we will find the values of D Discriminant and evaluate our answer accordingly .

Let us start with

$y = -3x^2 + x + 12\\a=-3 , b =1 , c =12\\D=1^2-4*(-3)*(12)\\D=1+144\\D=145\\D>0 \\$

Hence we have two real roots for this equation.

$y = 2x^2 - 6x + 5\\$

$y = 2x^2 - 6x + 5\\a=2,b=-6,c=5\\D=(-6)^2-4*2*5\\D=36-40\\D=-4\\D$

Hence we do not have any real root for this quadratic

$y = x^2 + 7x - 11\\a=1,b=7,-11\\D=7^2-4*1*(-11)\\D=49+44\\D=93\\$

Hence D>0 and thus we have two real roots for this equation.

$y = -x^2 - 8x - 16\\a=-1,b=-8,c=-16\\D=(-8)^2-4*(-1)*(-16)\\D=64-64\\D=0\\$

Hence we have one real root to this quadratic equation.

7 0

3 years ago

here are several types of objects. For each type of object, estimate how many there are in a stack that is 5 feet high. Be prepa

lorasvet [3.4K]

We need to estimate how tall is, on average, one of these object, and then count how many would be in a 5 feet high stack.

For example, on Google you may see that "1.5 cubic foot boxes are the standard box, manufactured by most companies". So, we assume that a standard cardboard box is 1.5 feet tall.

So, if we set the equation

$1.5k = 5 \iff k=\dfrac{5}{1.5}=3.\bar{3}$

So, there would be between 3 and 4 cardboard boxes in a 5 feet tall stack.

Similarly, we can see that the average book is 9 inches tall. 9 inches are 0.75

feet, so we have

$0.75k=5 \iff k=6.\bar{6}$

So, there would be between 6 and 7 books in a 5 feet tall stack.

The average brick is 75 millimeters tall, which means 0.25 feet tall. Again, we have

$0.25k=5 \iff k=20$

So, there would be 20 bricks in a 5 feet tall stack.

Finally, a coin is about 0.006 feet, which leads to

$0.006k=5 \iff k=833.\bar{3}$

So, there would be between 833 and 834 coins in a 5 feet tall stack.

5 0

3 years ago

For the rhombus below, find the measure of angles 1,2,3 and 4.

Alexus [3.1K]

Answer:

∠1 is 33°

∠2 is 57°

∠3 is 57°

∠4 is 33°

Step-by-step explanation:

First off, we already know that ∠2 is 57° because of alternate interior angles.

Second, it's important to know that rhombus' diagonals bisect each other; meaning they form 90° angles in the intersection. Another cool thing is that the diagonals bisect the existing angles in the rhombus. Therefore, 57° is just half of something.

Then, you basically just do some other pain-in-the-butt things after.

Since that ∠2 is just the bisected half from one existing angle, that means that ∠3 is just the other half; meaning that ∠3 is 57°, as well.

Next is to just find the missing angle ∠1. Since we already know ∠3 is 57°, we can just add that to the 90° that the diagonals formed at the intersection.

57° + 90° = 147°

180° - 147° = 33°

∠1 is 33°

Finally, since that ∠4 is just an alternate interior angle of ∠1, ∠4 is 33°, too.

5 0

3 years ago

Plz answer this question I need help

spin [16.1K]

This image is not loading therefore no one can answer it

7 0

3 years ago