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Norma-Jean [14]

3 years ago

10

Help me pls i dont have long i have 10 mins this is 1 of 2 questions can you pls answer both ill give brainly when 2 answers are

up (NO BOTS OR I WILL REPORT)

2 answers:

astraxan [27]3 years ago

8 0

I’ll help you with your math I am very good at it

Andreas93 [3]3 years ago

6 0

Message me for help! I’m a new toutor that would love to help you!

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Is 28/49=4/7 a ratio proportion

leva [86]

28 / 49
divide by 7 at top and bottom

(28/7) / (49/7)
4 / 7

5 0

3 years ago

Give P(-1, 8) and Q(5,12), and MN is the image of PQ under a translation of T

Brilliant_brown [7]

Coordinates of N are (-55, 6)

4 0

2 years ago

What is wrong with the following equation?x2+xâˆ’20/xâˆ’4=x+5a. (xâˆ’4)(xâˆ’5)â‰ x2+xâˆ’20b. The left-hand side is not defined f

joja [24]

Answer:

The answer to this question can be defined as follows:

In part (i), the answer is "option d".

In part (ii), the answer is "option 2".

Step-by-step explanation:

Given:

Part (a)

$\Rightarrow \bold{\frac{x^2+x-20}{x-4}=x+5}\\\\$

Solve the above equation:

$\Rightarrow x^2+x-20=(x+5)(x-4)\\\\\Rightarrow x^2+x-20=(x^2-4x+5x-20)\\\\\Rightarrow x^2+x-20=x^2-4x+5x-20\\\\\Rightarrow \boxed{x^2+x-20=x^2+x-20}\\$

Given:

Part (b)

$\Rightarrow \bold{ \lim_{x \to \3} \frac{x^2+x-12}{x-3}= \lim_{x \to 3} (x+4)}\\\\$

Solve the above equation:

factor of $\Rightarrow x^2+x-20 =(x-3)(x+4)$

$\Rightarrow \lim_{x \to \3} \frac{(x-3)(x+4)}{x-3}= \lim_{x \to 3} (x+4)\\\\\Rightarrow \lim_{x \to \3} (x+4)= \lim_{x \to 3} (x+4)\\\\$

apply limit value:

$\Rightarrow (3+4)= (3+4)\\\\\Rightarrow 7= 7$

6 0

3 years ago

How many distinct triangles can be formed for which m∠E = 64°, g = 9, and e = 10?<br> triangle(s)

Shalnov [3]

Using the law of sines, it is found that 1 triangle can be formed with the given conditions.

<h3>What is the law of sines?</h3>

Suppose we have a triangle in which:

The length of the side opposite to angle A is a.

The length of the side opposite to angle B is b.

The length of the side opposite to angle C is c.

The lengths and the sine of the angles are related as follows:

$\frac{\sin{A}}{a} = \frac{\sin{B}}{b} = \frac{\sin{C}}{c}$

Hence, for this problem, we have that the relation is:

$\frac{\sin{64^\circ}}{10} = \frac{\sin{G}}{9}$

$\sin{G} = 0.9\sin{64^\circ}$

$\sin{G} = 0.8089$

$G = \arcsin{0.8089}$

G = 54º

A triangle can have one angle of 64º and another of 54º, hence 1 triangle can be formed with the given conditions.

More can be learned about the law of sines at brainly.com/question/25535771

#SPJ1

5 0

2 years ago

Write the letter for each expression in the box under the correct category in the table.

Ad libitum [116K]

Answer:

a) 9+(-3) + (-2) = 9 -5 = 4> 0 ⇒ Positive Sum

b) -12 +1 +11 = -12 +12 5 = 0 ⇒ Zero Sum

c) -3 +8 + (-4) = -7 + 8 -5 =-1 < 0 ⇒ Negative Sum

d) -4+9+(-10) = 9 -14 = -5 < 0 ⇒ Negative Sum

e) 3+(-2) + (-4) = 3 -6= -3 < 0 ⇒ Negative Sum

f) 2 + (-4) + 2 = 4 -4 = 0 ⇒ Zero Sum

Step-by-step explanation:

Positive Sum is defined as the condition in which the Sum of total of gains and losses is GREATER THAN ZERO.

Zero Sum is defined as the condition in which the Sum of total of gains and losses is EQUAL TO ZERO.

Negative Sum is defined as the condition in which the Sum of total of gains and losses is LESS THAN ZERO.

Now, here

a) 9+(-3) + (-2) = 9 -5 = 4> 0 ⇒ Positive Sum

b) -12 +1 +11 = -12 +12 5 = 0 ⇒ Zero Sum

c) -3 +8 + (-4) = -7 + 8 -5 =-1 < 0 ⇒ Negative Sum

d) -4+9+(-10) = 9 -14 = -5 < 0 ⇒ Negative Sum

e) 3+(-2) + (-4) = 3 -6= -3 < 0 ⇒ Negative Sum

f) 2 + (-4) + 2 = 4 -4 = 0 ⇒ Zero Sum

5 0

2 years ago