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aalyn [17]
3 years ago
15

Please heeelp !! Missing triangles !!

Mathematics
2 answers:
Airida [17]3 years ago
6 0

Answer: ABC seem to be equilateral

so A=B=C=60

D=42

E=35

F=35

G=55

H=45

I=45

j=15

k=15

L=75

M=180

N=11

O=35

P=11

Q=55

R=55

S=32

T=32

U=69

Step-by-step explanation:

Ymorist [56]3 years ago
3 0

And the solution please!

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Divide the rational expressions and express in simplest form. When typing your answer for the numerator and denominator be sure
Veseljchak [2.6K]

Dividing by a fraction is equivalent to multiply by its reciprocal, then:

\begin{gathered} \frac{3y^2-7y-6}{2y^2-3y-9}\div\frac{y^2+y-2}{2y^2+y-3^{}}= \\ =\frac{3y^2-7y-6}{2y^2-3y-9}\cdot\frac{2y^2+y-3}{y^2+y-2}= \\ =\frac{(3y^2-7y-6)(2y^2+y-3)}{(2y^2-3y-9)(y^2+y-2)} \end{gathered}

Now, we need to express the quadratic polynomials using their roots, as follows:

ay^2+by+c=a(y-y_1)(y-y_2)

where y1 and y2 are the roots.

Applying the quadratic formula to the first polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{7\pm\sqrt[]{(-7)^2-4\cdot3\cdot(-6)}}{2\cdot3} \\ y_{1,2}=\frac{7\pm\sqrt[]{121}}{6} \\ y_1=\frac{7+11}{6}=3 \\ y_2=\frac{7-11}{6}=-\frac{2}{3} \end{gathered}

Applying the quadratic formula to the second polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot2\cdot(-3)}}{2\cdot2} \\ y_{1,2}=\frac{-1\pm\sqrt[]{25}}{4} \\ y_1=\frac{-1+5}{4}=1 \\ y_2=\frac{-1-5}{4}=-\frac{3}{2} \end{gathered}

Applying the quadratic formula to the third polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{3\pm\sqrt[]{(-3)^2-4\cdot2\cdot(-9)}}{2\cdot2} \\ y_{1,2}=\frac{3\pm\sqrt[]{81}}{4} \\ y_1=\frac{3+9}{4}=3 \\ y_2=\frac{3-9}{4}=-\frac{3}{2} \end{gathered}

Applying the quadratic formula to the fourth polynomial:

\begin{gathered} y_{1,2}=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ y_{1,2}=\frac{-1\pm\sqrt[]{1^2-4\cdot1\cdot(-2)}}{2\cdot1} \\ y_{1,2}=\frac{-1\pm\sqrt[]{9}}{2} \\ y_1=\frac{-1+3}{2}=1 \\ y_2=\frac{-1-3}{2}=-2 \end{gathered}

Substituting into the rational expression and simplifying:

\begin{gathered} \frac{3(y-3)(y+\frac{2}{3})2(y-1)(y+\frac{3}{2})}{2(y-3)(y+\frac{3}{2})(y-1)(y+2)}= \\ =\frac{3(y+\frac{2}{3})}{2(y+2)}= \\ =\frac{3y+2}{2y+4} \end{gathered}

8 0
1 year ago
Caleb bought groceries and paid $1.60 in sales tax. The sales tax rate is 2.590.
slava [35]
Good for Caleb, he better have bought cheese.
6 0
2 years ago
The height of a triangle is 4 m less than its base. The area of the triangle is 48 m².What is the length of the base?
yaroslaw [1]

Triangle Area = .5 * base * height

48 = .5 * (x +4) * x

48 = (.5x + 2) * x

48 = .5x^2 + 2x

.5x^2 + 2x -48 = 0

x = 8

base = 12

5 0
3 years ago
Read 2 more answers
A donut store has 11 different types of donuts. You can only buy a bag of 3 of them, where each donut has to be of a different t
MakcuM [25]

Answer:

165.

Step-by-step explanation:

Since repetition isn't allowed, there would be 11 choices for the first donut, (11 - 1) = 10 choices for the second donut, and (11 - 2) = 9 choices for the third donut. If the order in which donuts are placed in the bag matters, there would be 11 \times 10 \times 9 unique ways to choose a bag of these donuts.

In practice, donuts in the bag are mixed, and the ordering of donuts doesn't matter. The same way of counting would then count every possible mix of three donuts type 3 \times 2 \times 1 = 6 times.

For example, if a bag includes donut of type x, y, and z, the count 11 \times 10 \times 9 would include the following 3 \times 2 \times 1 arrangements:

  • xyz.
  • xzy.
  • yxz.
  • yzx.
  • zxy.
  • zyx.

Thus, when the order of donuts in the bag doesn't matter, it would be necessary to divide the count 11 \times 10 \times 9 by 3 \times 2 \times 1 = 6 to find the actual number of donut combinations:

\begin{aligned} \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}.

Using combinatorics notations, the answer to this question is the same as the number of ways to choose an unordered set of 3 objects from a set of 11 distinct objects:

\begin{aligned}\begin{pmatrix}11 \\ 3\end{pmatrix} &= \frac{11 !}{(11 - 3)! \times 3 !} \\ &= \frac{11 !}{8 ! \times 3 !} \\ &= \frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\end{aligned}.

5 0
2 years ago
Use the integer tiles to evaluate the following expressions. 6 + 3 = 6 + (–4) = 6 + (–6) =
babunello [35]

Given:

The expressions are:

6+3

6+(-4)

6+(-6)

To find:

The value of given expression by using integer tiles.

Solution:

We have,

6+3

Here, both number are positive. When we add 6 and 3 positive integer tiles, we get 9 positive integer tiles as shown in the below figure. So,

6+3=9

Similarly,

6+(-4)

Here, 6 is positive and -4 is negative. It means we have 6 positive integer tiles and 4 negative integer tiles.

When we cancel the positive and negative integer tiles, we get 2 positive integer tiles as shown in the below figure. So,

6+(-4)=2

6+(-6)

Here, 6 is positive and -6 is negative. It means we have 6 positive integer tiles and 6 negative integer tiles.

When we cancel the positive and negative integer tiles, we get 0 integer tiles as shown in the below figure. So,

6+(-6)=0

Therefore, 6+3=9, 6+(-4)=2,6+(-6)=0.

8 0
2 years ago
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