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2 years ago

I'm kinda lost in this problem. Can you teach me how to solve these

Mathematics

1 answer:

boyakko [2]2 years ago

8 0

53) Vertical angles are congruent, so $m\angle IDA=121^{\circ}$

54) $\angle YDA$ and $\angle YDF$ are supplementary, so $m\angle YDA=180^{\circ}-121^{\circ}=59^{\circ}$

55) Because vertical angles are congruent, $m\angle FDI=59^{\circ}$ . Now, because DR bisects $\angle FDI$ , this means $m\angle RDI=\frac{59}{2}=29.5^{\circ}$

56) This is a geometric sequence where the next term is obtained by multiplying the previous term by -2, so the next two numbers are 16, -32

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10×10 by the power of 2 equal

ryzh [129]

Answer:

10,000

Step-by-step explanation:

10 x 10 = 100 when you do the power of 2, you multiply that number by itself.

7 0

3 years ago

Use fraction strips to help you complete the number line then locate and draw a point for the fraction 1/3

Finger [1]

<span>Halves, thirds, fourths, sixths, eighths (e.g., 1/2, 1/3, 1/4, 1/6, 1/8) these are some answers </span>

6 0

3 years ago

Area and circumference of this circle

faust18 [17]

Answer:

Step-by-step explanation:

Given: The radius is 6 cm

To find the area + circumference we need to use 2 formulas:

Area: pi* radius^2

Circumference: 2*pi*r

First we can do the area

I will use "pi" for pi instead of 3.14

pi * 6^2

= 36 pi

The area is 36 pi

Next, circumference

2 * pi *r

2*pi*6

= 12 pi

So the area is 36 pi, and the circumference is 12 pi

5 0

3 years ago

Solve these linear equations by Cramer's Rules Xj=det Bj / det A:

timurjin [86]

Answer:

(a) $x_1=-2,x_2=1$

(b) $x_1=\frac{3}{4} ,x_2=-\frac{1}{2} ,x_3=\frac{1}{4}$

Step-by-step explanation:

(a) For using Cramer's rule you need to find matrix $A$ and the matrix $B_j$ for each variable. The matrix $A$ is formed with the coefficients of the variables in the system. The first step is to accommodate the equations, one under the other, to get $A$ more easily.

$2x_1+5x_2=1\\x_1+4x_2=2$

$\therefore A=\left[\begin{array}{cc}2&5\\1&4\end{array}\right]$

To get $B_1$ , replace in the matrix A the 1st column with the results of the equations:

$B_1=\left[\begin{array}{cc}1&5\\2&4\end{array}\right]$

To get $B_2$ , replace in the matrix A the 2nd column with the results of the equations:

$B_2=\left[\begin{array}{cc}2&1\\1&2\end{array}\right]$

Apply the rule to solve $x_1$ :

$x_1=\frac{det\left(\begin{array}{cc}1&5\\2&4\end{array}\right)}{det\left(\begin{array}{cc}2&5\\1&4\end{array}\right)} =\frac{(1)(4)-(2)(5)}{(2)(4)-(1)(5)} =\frac{4-10}{8-5}=\frac{-6}{3}=-2\\x_1=-2$

In the case of B2, the determinant is going to be zero. Instead of using the rule, substitute the values of the variable $x_1$ in one of the equations and solve for $x_2$ :

$2x_1+5x_2=1\\2(-2)+5x_2=1\\-4+5x_2=1\\5x_2=1+4\\ 5x_2=5\\x_2=1$

(b) In this system, follow the same steps,ust remember $B_3$ is formed by replacing the 3rd column of A with the results of the equations:

$2x_1+x_2 =1\\x_1+2x_2+x_3=0\\x_2+2x_3=0$

$\therefore A=\left[\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right]$

$B_1=\left[\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right]$

$B_2=\left[\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right]$

$B_3=\left[\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right]$

$x_1=\frac{det\left(\begin{array}{ccc}1&1&0\\0&2&1\\0&1&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{1(2)(2)+(0)(1)(0)+(0)(1)(1)-(1)(1)(1)-(0)(1)(2)-(0)(2)(0)}{(2)(2)(2)+(1)(1)(0)+(0)(1)(1)-(2)(1)(1)-(1)(1)(2)-(0)(2)(0)}\\ x_1=\frac{4+0+0-1-0-0}{8+0+0-2-2-0} =\frac{3}{4} \\x_1=\frac{3}{4}$

$x_2=\frac{det\left(\begin{array}{ccc}2&1&0\\1&0&1\\0&0&2\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)} =\frac{(2)(0)(2)+(1)(0)(0)+(0)(1)(1)-(2)(0)(1)-(1)(1)(2)-(0)(0)(0)}{4} \\x_2=\frac{0+0+0-0-2-0}{4}=\frac{-2}{4}=-\frac{1}{2}\\x_2=-\frac{1}{2}$

$x_3=\frac{det\left(\begin{array}{ccc}2&1&1\\1&2&0\\0&1&0\end{array}\right)}{det\left(\begin{array}{ccc}2&1&0\\1&2&1\\0&1&2\end{array}\right)}=\frac{(2)(2)(0)+(1)(1)(1)+(0)(1)(0)-(2)(1)(0)-(1)(1)(0)-(0)(2)(1)}{4} \\x_3=\frac{0+1+0-0-0-0}{4}=\frac{1}{4}\\x_3=\frac{1}{4}$

6 0

3 years ago

The highest scores ata gymnastics meet were 9.675, 9.25,9.325 and 9.5. Write the scores in order from least to greatest.

Inga [223]

Starting with the least score it would be ; 9.25 , 9.325 , 9.5 , 9.675

4 0

3 years ago

Read 2 more answers

I'm kinda lost in this problem. Can you teach me how to solve these ​

I'm kinda lost in this problem. Can you teach me how to solve these