HELP QUICK PLEASE!

Which postulate, if any, could be used to prove the triangles are congruent?
Hypotenuse-Leg (HL)
Side-Angle-Side (SAS)
O Side-Side-Side (SSS)
The triangles cannot be proved congruent.

Answer:

12

Step-by-step explanation:

for this you need to draw a straight line and plot the points J,K,L,M,N according to the question.

here,

given K lies between J and M, L is between K and M , M is between K and N.

JN= 30, KM=8

JK=KL=LM

MJ= ?

according to the question,

J--------K-------L--------M-------N

KM= 8

KL+LM= 8 (KL+LM=KM)

KL+KL=8 ( given, LM=KL)

2KL= 8

KL=4

so,

JK= KL=4 and LM= KL= 4

now,

MJ= ?

we know

MJ= JK+KM = 4+8= 12

Answer:

10/3 days  or 3 1/3 days

Step-by-step explanation:

Let d = number of days total

75% of the days is 10

.75 d = 10

d = 10/.75

d = 40/3 days

We want to know how man is 25% of d

d * 25%

(40/3) *.25

10/3 days

Step-by-step explanation:

Hi there!

Please see the answer in the picture.

Hope it helps!

1. Approach

One is given a trigonometric equation with and one is asked to prove that it is true. Using the attached image, combined with the knowledge of trigonometry, one can evaluate each trigonometric function. Then one can simplify each ratio to solve. To yield the most accurate result, one has to each of the ratios in a fractional form, rather than simplifying it into a decimal form. Remember the right angle trigonometric ratios, these ratios describe the relationship between the sides and angles in a right triangle. Such ratios are as follows,

[tex]sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}\\\\csc(\theta)=\frac{hypotenuse}{opposite}\\\\sec(\theta)=\frac{hypotenuse}{adjacent}\\\\cot(\theta)=\frac{adjacent}{opposite}[/tex]

Please note that the terms (opposite) and (adjacent) are relative to the angle uses in the ratio, however the term (hypotenuse) refers to the side opposite the right angle, this side never changes its name. Use these ratios to evaluate the trigonometric functions. Then simplify to prove the identity.

2. Problem (9)

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

As per the attached image, the following statements regarding the value of each ratio can be made:

[tex]sin(60)=\frac{\sqrt{3}}{2}\\\\cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cos(60)=\frac{1}{2}[/tex]

Substitute,

[tex]\frac{sin(60)+cos(30)}{1+sin(30)+cos(60)}=sin(60)[/tex]

[tex]\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

Simplify,

[tex]\frac{\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\frac{2\sqrt{3}}{2}}{1+\frac{1}{2}+\frac{1}{2}}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\frac{2\sqrt{3}}{2}}{1+1}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\sqrt{3}}{1+1}=\frac{\sqrt{3}}{2}[/tex]

[tex]\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}[/tex]

Thus, this equation is true.

2. Problem (10)

Use a similar strategy to evaluate this equation,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

Use the attached image to evaluate the ratios.

[tex]cos(30)=\frac{\sqrt{3}}{2}\\\\sin(30)=\frac{1}{2}\\\\cot(60)=\frac{1}{\sqrt{3}}[/tex]

Substitute,

[tex]\frac{1-cos(30)}{sin(30)}=\frac{1-cot(60)}{1+cot(60)}[/tex]

[tex]\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

Simplify,

[tex]\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{1-\frac{1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{1+\frac{1}{\sqrt{3}}}[/tex]

[tex]\frac{\frac{2-\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}[/tex]

[tex]2-\sqrt{3}=\frac{\frac{\sqrt{3}-1}{\sqrt{3}}}{\frac{\sqrt{3}+1}{\sqrt{3}}}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}+1}}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}[/tex]

Rationalize the denominator,

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}[/tex]

[tex]2-\sqrt{3}=\frac{\sqrt{3}-1}{\sqrt{3}+1}*\frac{\sqrt{3}-1}{\sqrt{3}-1}[/tex]

[tex]2-\sqrt{3}=\frac{(\sqrt{3}-1)^2}{3-1}[/tex]

[tex]2-\sqrt{3}=\frac{3-2\sqrt{3}+1}{2}[/tex]

[tex]2-\sqrt{3}=\frac{4-2\sqrt{3}}{2}[/tex]

[tex]2-\sqrt{3}=2-\sqrt{3}[/tex]

Therefore, this equation is also true.

Answer:

7.20 dollars

Step-by-step explanation:

1 1/4 can be expressed as 5/4 ((1*4+1) /4)

If price of 5/4 pound of turkey = $9

Price of 1 pound of turkey = $9 / (5/4)

( Same way you'd divide the price of 2 pounds by 2 to get the price of 1 pound)

$9 / (5/4) = $7.2

the answer is B. 7

Step-by-step explanation:

it's the number u see the most therefore its the most common

Answer:

D. 10 maybe?

Step-by-step explanation:

because it moves a decimal with every number. im not sure I hope it's correct. it adds on a zero

9514 1404 393

Answer:

Step-by-step explanation:

Substitute the expressions.

After the first translation, the value of x is (x+n). Put that as the value of x in the second translation.

  x ⇒ x +s . . . . . . . . . the definition of the second translation

  (x+n) ⇒ (x+n) +s . . . the result after both translations

The same thing goes for y. After the first translation, its new value is (y+1).

  y ⇒ y +m . . . . . . . . the definition of the second translation

  (y+1) ⇒ (y+1) +m . . . the result of both translations

Then the composition of A followed by B is (x, y) ⇒ (x +n +s, y +1 +m).

__

The same reasoning applies. After the B translation, the x-coordinate is (x+s) and the y-coordinate is (y+m). Then the A translation changes these to ...

  x ⇒ x +n . . . . . . . . . . definition of translation A

  (x+s) ⇒ (x+s) +n . . . . translation A operating on point translated by B

  y ⇒ y +1 . . . . . . . . . . . definition of translation A

  (y+m) ⇒ (y+m) +1 . . . . translation A operating on point translated by B

The composition of B followed by A is (x, y) ⇒ (x + s + n, y + m + 1).

__

You should recognize that the sums (x+n+s) and (x+s+n) are identical. The commutative and associative properties of addition let us rearrange the order of the terms with no effect on the outcome.

The two translations give the same result in either order.

9514 1404 393

Answer:

  2. (only)

Step-by-step explanation:

The Pythagorean theorem tells you the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse. To determine if these are right triangles, determine if that condition is met.

  1. 3^2 +5^2 = 9 + 25 = 34 ≠ (√35)^2 . . . . not a right triangle

  2. 5^2 +4^2 = 25 +16 = 41 = (√41)^2 . . . a right triangle

  3. 6^2 +8^2 = 36 +64 = 100 ≠ (√10)^2 . . . . not a right triangle

  4. 3^2 +3^2 = 9 +9 = 18 ≠ (3√3)^2 = 27 . . . . not a right triangle

Answer:

45

Step-by-step explanation:

CHECK ANSWER:

45  *5 = 225

45+5 = 50

50 < 225

sum < times

I hope this helped !

Answer:

Infinite solutions

Step-by-step explanation:

x + 4y = 7 --------------(I)

8y = 14 - 2x

⇒ 2x + 8y = 14 -----------(II)

[tex]\frac{a_{1}}{a_{2}}=\frac{1}{2}\\\\\frac{b_{1}}{b_{2}}=\frac{4}{8}=\frac{1}{2}\\\\\frac{c_{1}}{c_{2}}=\frac{7}{14}=\frac{1}{2}\\\\\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{1}{2}[/tex]

So, infinite solutions

Answer:

G

Step-by-step explanation:

Given the 2 equations

x + 4y = 7 → (1)

8y = 14 - 2x → (2)

make x the subject of (1) by subtracting 4y from both sides

x = 7 - 4y

Substitute x = 7 - 4y into (2)

8y = 14 - 2(7 - 4y) ← distribute

8y = 14 - 14 + 8y

8y = 8y

Since both sides are equal this indicates the system has infinite solutions

Answer:

10cm

Step-by-step explanation:

Assuming the shape is a rectangle since it's already stated that it's a parallelogram, and the area is stated, we can use the Pythagorean theorem to find the length

[tex]a^{2} +b^2=c^2[/tex], where c will be the length

isolate c

[tex]c^2=a^2+b^2[/tex]

[tex]c=\sqrt{a^2+b^2}[/tex]

substitute for a and b

[tex]c=\sqrt{8^2+6^2}[/tex]

[tex]c=\sqrt{64+36}[/tex]

[tex]c=\sqrt{100}[/tex]

[tex]c=10[/tex]

Step-by-step explanationIf you are reading this say

thank u