LiIIiam0216

There are 4 ways.

Since △DEF is equilateral we know DE=DF=8.

Since AC⊥DE we know △ACE is a right triangle.

We are given that CE=4 and want to find AE.

Using the Pythagorean Theorem on △ACE we have AE^2 = AC^2 − CE^2.

To find AC we use the Pythagorean Theorem on △ACD where AD = DE − DC = 8 − 5 = 3 : AC^2 = AD^2 + CD^2 = 3^2 + 5^2 =34.

Thus, AE^2 = 34 − 4^2 = 26.

Taking the positive square root of both sides gives AE = sqrt(26)​.

Since the quadratic equation has real coefficients and the roots are complex numbers, the roots come in complex conjugate pairs. This means that the other root must be 5−3i.

We can use the fact that the sum of the roots of a quadratic equation is equal to the negative of the coefficient of our z term. In other words, the sum of the roots is b. So, we have:

$$ b = 5 + 3i + (5 - 3i) = 10. $$

We can also use the fact that the product of the roots of a quadratic equation is equal to the constant term. In other words, the product of the roots is c. So, we have:

$$ c = (5 + 3i)(5 - 3i) = 5^2 - (3i)^2 = 25 + 9 = 34. $$

Therefore, b + c = 10 + 34 = 44​.

tan (2 theta) = 3/4

2 theta = 0.932

theta = 0.466

The greatest common divisor of 957 and 1537 is 87.

Let's denote each two-digit number as $10a + b,$ where $a$ is the tens digit and $b$ is the units digit. Let $n$ have $k$ two-digit numbers in its decomposition.

When we calculate the alternating sum of these two-digit numbers, we get $10a_1 + b_1 - 10a_2 - b_2 + 10a_3 + b_3 - \cdots \pm 10a_k + b_k,$ where $a_i$ is the tens digit of the $i$-th two-digit number and $b_i$ is the units digit.

This can be rewritten as $10(a_1 - a_2 + a_3 - \cdots \pm a_k) + (b_1 - b_2 + b_3 - \cdots \pm b_k).$ Let $A = a_1 - a_2 + a_3 - \cdots \pm a_k$ and $B = b_1 - b_2 + b_3 - \cdots \pm b_k.$ Then, the sum becomes $10A + B.$

Since $10A + B$ is divisible by $m,$ it implies $10A + B \equiv 0 \pmod{m},$ which gives $10A \equiv -B \pmod{m}.$

We know that $10 \equiv 1 \pmod{m},$ so $10A \equiv A \pmod{m}.$ Hence, we have $A \equiv -B \pmod{m}.$

Therefore, $n$ is divisible by $m$ if and only if $10a + b$ is divisible by $m,$ implying that the divisibility test works for $m = \boxed{11}.$

We know that a quadratic function with roots at x=2 and x=4 can be written in the factored form as:

a(x−2)(x−4)

We are also given that the function takes the value 6 when x=3. This translates to the equation: a(3−2)(3−4)=6 which simplifies to −a=6

Since we already know that a cannot be 0 (the function wouldn't be quadratic), we can divide both sides by −1 to get a=−6

Plugging this value of a back into the factored form, we get the quadratic function: −6(x−2)(x−4)

Expanding this expression, we get the answer in the desired form: −6x^2 + 36x + 48

So the answer is -6x^2 + 36x + 48.

Nevermind, I solved this.

I solved this problem!  The answer is 100*sqrt(2).