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3 years ago

15

I am a factor of 16 but not a multiple of 4. Who am i?

1 answer:

swat323 years ago

6 0

1 & 2 are the only factors of 16 & not multiples of 4.

Hope this helps!!

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How many turning points can a polynomial with a degree of 7 have?

Gre4nikov [31]

The best answer to this question is A because the graph of a polynomial can have up to 1 less turning point than its highest degree. For example, a 10th degree polynomial can have 9 or less turning points.

6 0

3 years ago

HELP PLEASE AND THANK YOU 1JAAB

Dafna1 [17]

Answer:

$\bold{Q1}\ side\ BD\\\\\bold{Q2}\ \angle5\ \leftrightarrow\ \angle6\\\\\bold{Q3}\ \triangle TAP\\\\\bold{Q4}\ \triangle APT$

Step-by-step explanation:

For Q1 and Q2

$\text{If}\ AB\cong AC,\ BD\cong CD,\ RSTU\ \text{is a parallelogram (rectangle)},\ \text{then}\\\\\angle 1\ \leftrightarrow\ \angle2\\\angle3\ \leftrightarrow\ \angle4\\\angle C\leftrightarrow\ \angle B\\\angle 5\ \leftrightarrow\ \angle6\\\angle7\ \leftrightarrow\ \angle 8\\\angle R\ \leftrightarrow\ \angle T\\\angle S\ \leftrightarrow\ \angle U$

$\overline{AB}\ \leftrightarrow\ \overline{AC}\\\overline{BD}\ \leftrightarrow\ \overline{CD}\\\\\overline{RS}\ \leftrightarrow\ \overline{TU}\\\overline{ST}\ \leftrightarrow\ \overline{UR}$

For Q3 and Q4

$O\ \leftrightarrow\ T\\B\ \leftrightarrow\ A\\R\ \leftrightarrow\ P\\\\E\ \leftrightarrow\ X\\F\ \leftrightarrow\ W\\D\ \leftrightarrow\ T$

4 0

3 years ago

The number of points scored in a basketball game by Slim Dunkin is shown in

koban [17]

The correct answer is C. Goodluck

7 0

3 years ago

Read 2 more answers

Explain why a quadratic equation with a positive discriminant has two real solutions, a quadratic equation with a negative discr

Bad White [126]

Answer:

A quadratic equation can be written as:

a*x^2 + b*x + c = 0.

where a, b and c are real numbers.

The solutions of this equation can be found by the equation:

$x = \frac{-b +- \sqrt{b^2 - 4*a*c} }{2*a}$

Where the determinant is D = b^2 - 4*a*c.

Now, if D>0

we have the square root of a positive number, which will be equal to a real number.

√D = R

then the solutions are:

$x = \frac{-b +- R }{2*a}$

Where each sign of R is a different solution for the equation.

If D< 0, we have the square root of a negative number, then we have a complex component:

√D = i*R

$x = \frac{-b +- C*i }{2*a}$

We have two complex solutions.

If D = 0

√0 = 0

then:

$x = \frac{-b +- 0}{2*a} = \frac{-b}{2a}$

We have only one real solution (or two equal solutions, depending on how you see it)

3 0

3 years ago

Help! This is so confusing please

Rus_ich [418]

Given equation is $f(x)=6^x$

Now it says to find f(2)

that simply means plug x=2 because we see that 2 is written in place of x in f(x) and f(2)

$f(2)=6^2$

$f(2)=6*6$

$f(2)=36$

Hence final answer is $f(2)=36$

4 0

4 years ago

Read 2 more answers