For the following equation, find a solution for the variable. Show all of your work and use complete sentences to explain the solving process that you used to find a solution for the equation. Be sure to include at least two terms from the word bank. 4+ x = 7

So basically what you have to do is first identify all of the sides or angles that equal eachother then you can see if you did a side side and side in a row etc, like 44 would be SAS because the vertical angles would be inbetween the two sides. 46 would be SSS because after you make the line in the middle for reflexive property you can see that there are three sides in a row. If this didnt help i recommend looking on Khan Academy

Answer:

the answer is number 1 !!

• correct me if I'm wrong

Answer: If angle A is congruent to angle E

Answer:

true

Step-by-step explanation:

Answer:

10.5

Step-by-step explanation:

A=a+b

2h=2+5

2·3=10.5

Answer:

solution given:

let's see only in a right-angled triangle Δ ACD.

AC=3 units

AD=5 units

since Δ ACD is a right-angled triangle. It satisfies Pythagoras law

[tex]AD^{2}=AC^{2}+CD^{2}[/tex]

25=9+[tex]CD^2[/tex]

[tex]CD^2[/tex]=25-9=16

[tex]CD=\sqrt{16} =4 units[/tex]

Now

Area of rectangle Δ ACD=[tex]\frac{1}{2} CD*AC=\frac{1}{2}*4*3=6\: square \:units[/tex]

similarly,

[tex]\frac{1}{2} AD*CE=6\: square \:units\\ 5*CE=6*2\\CE=\frac{12}{5}=2.4 units[/tex]

since AB=CE=2.4 units

Δ ACE is a right-angled triangle. It satisfies Pythagoras law.

[tex]AC^{2}=AE^{2}+CE^{2}\\3^2=AE^2+2.4^2\\9-5.76=AE^2\\AE and BC=\sqrt{1.24} =1.11 units[/tex]

now

area of trapezoid=area ofΔACD+Area of ΔABC

=6+[tex]\frac{1}{2}AB*BC=6+\frac{1}{2}*2.4*1.11=6+1.33 =7.33\: square \:units[/tex]

Step-by-step explanation:

Answer:

70% right, 30% wrong

Step-by-step explanation:

3 is 30% of 10

Answer:

gek

Step-by-step explanation:

Answer:

292.6

Step-by-step explanation:

Answer:

55.56%

Step-by-step explanation:

55/99=.5555555...

.55555...*100=55.555...%

55.56%

Answer:

A. 511; 6

Step-by-step explanation:

The margin of error would be [tex]\pm\frac{517-505}{2}=\pm\frac{12}{2}=\pm 6[/tex]

The sample mean would be [tex]505+6=511[/tex] (or [tex]517-6=511[/tex])

Answer:

I feel like something is wrong with the equation but the answer is a = 3

Step-by-step explanation:

a + 3a - 3 = 6 + 3 = 9

--> 4a = 12 --> a = 3

Answer:

Step-by-step explanation:

Drake has 2 houses and makes $50,000,000

Answer: store b because you can get more ground beat.

Step-by-step explanation:

Answer:

-5.

Step-by-step explanation:

5 + -5 = 0, so i's -5.

Answer:

this is do confusing I do not understand

150−28w

i think i dont know

Answer:

the perimeter of the figure is 27.4 mi

Step-by-step explanation:

rectangle perimeter = 6+6+6 = 18

half circle perimeter = 1/2 2πr = 3.14 × 3 = 9.42

the perimeter of the figure is 18+9.42 = 27.42 --> 27.4 mi

Answer:

D: {-2, 0, 2}

R: {-1, 1, 3

Step-by-step explanation:

Two t-shirts that are(31.80) plus a hat(6.20) cost £38.00.

which means that one t-shirt costs £15.90 :)

Answer:

D

Step-by-step explanation:

Pluck in the values, we have:

4.5/2.4 - 6 = 20/2 = 10

Answer:

Step-by-step explanation:

[tex]x - 0.20x \: = 1480[/tex]

[tex]0.8x = 1480[/tex]

[tex]x = 1480 \div 0.8[/tex]

I'm not sure if this is correct, but I got this

Last year’s computer models are discounted 20%. If a computer is on sale for $1,480, what is the original price of the computer rounded to the nearest dollar?

20% = 0.2

1,480 x 0.2 = 296

1,480 + 296 = 1,776

Answer: 1,776

Again, I don't know if this is correct, so don't be 100% sure with my response.

Answer:

1/6

Step-by-step explanation:

P(A) = probability a student buys lunch = 60% = 0.6

P(B) = probability a student buys lunch and dessert = 10% = 0.1

We are looking at the probability that a student buys lunch and dessert on the condition that they already bought lunch. Therefore, we are looking at a conditional probability.

P(B given A) = P(A and B) / P(A)

= 0.1 / 0.6

= 1/6 ≈ 0.167

Answer:

D. [tex]-5 \frac{5}{6}[/tex]

Step-by-step explanation:

Add the values:

[tex]-3\frac{2}{3}+\left(-2\right)\frac{1}{6}[/tex]

[tex]=-\frac{11}{3}+\left(-2\right)\frac{1}{6}[/tex]

[tex]=-\frac{11}{3}-\frac{1}{3}[/tex]

[tex]-5 \frac{5}{6}[/tex]

Answer:

C. - 1 ½

Step-by-step explanation:

-3⅔ - x = - 2⅙

x = -3⅔ + 2⅙

x = -11/3 + 13/6

x = -22/6 + 13/6

x = -9/6

x = -1 ½

a) If f(y) is a probability density function, then both f(y) ≥ 0 for all y in its support, and the integral of f(y) over its entire support should be 1. eˣ > 0 for all real x, so the first condition is met. We have

[tex]\displaystyle \int_{-\infty}^\infty f(y) \, dy = \frac14 \int_0^\infty e^{-\frac y4} \, dy = -\left(\lim_{y\to\infty}e^{-\frac y4} - e^0\right) = \boxed{1}[/tex]

so both conditions are met and f(y) is indeed a PDF.

b) The probability P(Y > 4) is given by the integral,

[tex]\displaystyle \int_{-\infty}^4 f(y) \, dy = \frac14 \int_0^4 e^{-\frac y4} \, dy = -\left(e^{-1} - e^0\right) = \frac{e - 1}{e} \approx \boxed{0.632}[/tex]

c) The mean is given by the integral,

[tex]\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \int_0^\infty y e^{-\frac y4} \, dy[/tex]

Integrate by parts, with

[tex]u = y \implies du = dy[/tex]

[tex]dv = e^{-\frac y4} \, dy \implies v = -4 e^{-\frac y4}[/tex]

Then

[tex]\displaystyle \int_{-\infty}^\infty y f(y) \, dy = \frac14 \left(\left(\lim_{y\to\infty}\left(-4y e^{-\frac y4}\right) - \left(-4\cdot0\cdot e^0\right)\right) + 4 \int_0^\infty e^{-\frac y4} \, dy\right)[/tex]

[tex]\displaystyle \cdots = \int_0^\infty e^{-\frac y4} \, dy[/tex]

[tex]\displaystyle \cdots = -4 \left(\lim_{y\to\infty} e^{-\frac y4} - e^0\right) = \boxed{4}[/tex]