All categories

3 years ago

Question is in the picture help I’m desperate

Mathematics

2 answers:

hichkok12 [17]3 years ago

8 0

Answer:

the measure of <BAC = 60° and

46) option D is not true

Elenna [48]3 years ago

4 0

Question 45

The two remote interior angles 3x and 2x add to the exterior angle 150. This is an application of the remote interior angle theorem

2x+3x = 150

5x = 150

x = 150/5

x = 30

angle BAC = 2x = 2*30 = 60 degrees

<h3>Answer: 60 degrees</h3>

===================================================

Question 46

Choice A is true because of what I mentioned in the previous problem.

Choice B is true because we can find that angle C is 180-150 = 30. Then we can say A+B+C = 2x+3x+30 = 180 as all three interior angles of any triangle always add to 180.

Choice C is true since the equation above x = 30 can be written as 3x-2x = 30. It's a bit of a strange thing to do this, but it's still a valid statement.

Choice D is not true since 2x+3x is 150, which isn't 180 or larger. Even if we didn't know about the 150 at all, we still can't have 2 interior angles of a triangle add to 180 or larger (refer back to choice B).

<h3>Answer: Choice D</h3>

You might be interested in

(PLEASE HELP DUE NOW! I DON'T UNDERSTAND!)

BlackZzzverrR [31]

Answer:

q 1 is 24.5 and q3 is 34

Step-by-step explanation:

bye my luvvv

6 0

3 years ago

Read 2 more answers

Please could you find the answers to the questions in the attachment.

Fudgin [204]

( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ )we need 3 equations
1. midpoint equation which is ( $\frac{x1+x2}{2} , \frac{y1+y2}{2}$ ) when you have 2 points

2. distance formula which is D= $\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

3. area of trapezoid formula whhic is (b1+b2) times 1/2 times height

so

x is midpoint of B and C
B=11,10
c=19,6
x1=11
y1=10
x2=19
y2=6
midpoint=( $\frac{11+19}{2} , \frac{10+6}{2}$ )
midpoint=( $\frac{30}{2} , \frac{16}{2}$ )
midpoint= (15,8)

point x=(15,8)

y is midpoint of A and D
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
midpoint=( $\frac{5+21}{2} , \frac{8+0}{2}$ )
midpoint=( $\frac{26}{2} , \frac{8}{2}$ )
midpoint=(13,4)

Y=(13,4)

legnths of BC and XY
B=(11,10)
C=(19,6)
x1=11
y1=10
x2=19
y2=6
D= $\sqrt{(19-11)^{2}+(6-10)^{2}}$
D= $\sqrt{(8)^{2}+(-4)^{2}}$
D= $\sqrt{64+16}$
D= $\sqrt{80}$
D= $4 \sqrt{5}$
BC= $4 \sqrt{5}$

X=15,8
Y=(13,4)
x1=15
y1=8
x2=13
y2=4
D= $\sqrt{(13-15)^{2}+(4-8)^{2}}$
D= $\sqrt{(-2)^{2}+(-4)^{2}}$
D= $\sqrt{4+16}$
D= $\sqrt{20}$
D= $2 \sqrt{5}$
XY= $2 \sqrt{5}$

the thingummy is a trapezoid
we need to find AD and BC and XY
we already know that BC= $4 \sqrt{5}$ and XY= $2 \sqrt{5}$

AD distance
A=5,8
D=21,0
x1=5
y1=8
x2=21
y2=0
D= $\sqrt{(21-5)^{2}+(0-8)^{2}}$
D= $\sqrt{(16)^{2}+(-8)^{2}}$
D= $\sqrt{256+64}$
D= $\sqrt{320}$
D= $4 \sqrt{2}$
AD= $4 \sqrt{2}$

so we have
AD= $4 \sqrt{2}$
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$

AD and BC are base1 and base 2
XY=height
so
(b1+b2) times 1/2 times height
( $4 \sqrt{2}+4 \sqrt{5}$ ) times 1/2 times $2 \sqrt{5}$ =
( $4 \sqrt{2}+4 \sqrt{5}$ ) times \sqrt{5} [/tex] =
$4 \sqrt{10}+4*5$ = $4 \sqrt{10}+20$ =80 $\sqrt{10}$ =252.982

X=(15,8)
Y=(13,4)
BC= $4 \sqrt{5}$
XY= $2 \sqrt{5}$
Area=80 $\sqrt{10}$ square unit or 252.982 square units

7 0

3 years ago

Here goes another question lol oops

Murrr4er [49]

Student 1 and 3.
$- (5r - 3s + 1) = - 5r + 3s - 1$
Move the negative into the brackets,
positive becomes negative while two negatives make a positive.

3 0

4 years ago

Read 2 more answers

Answer this problem<br><br> x- 2 = -3x + 10

Semmy [17]

The correct answer is x = 3

6 0

3 years ago

Read 2 more answers

The British Queen is to receive 4 foreign dignitaries. The digni- taries are sensitive to the order in which they are received.

Readme [11.4K]

Answer:

Number of Ways = ₄P₄ = 24

Step-by-step explanation:

Given that there are going to be 4 dignitaries, and that they are sensitive to the order (i.e the order of the dignitaries matter), hence the total number of ways they can be arranged can be found by permutating 4 dignitaries.

i.e

Number of Ways = ₄P₄ = 24

4 0

4 years ago