All categories

2 years ago

PLS HELP

Mathematics

1 answer:

Charra [1.4K]2 years ago

4 0

Answer: -418, 728, 446, -429, 1159, -696

Step-by-step explanation:

1. $n = -199 - 219 = \boxed{-418}$ .

2. $m = 438 - (-290) = 438 + 290 = \boxed{728}$ .

3. $128 - (-318) = 128 + 318 = \boxed{446}$ .

4. The sum of 429 and 429 is $429+429 = 858$ . We need to subtract this from 429, so $429 - 858 = \boxed{-429}$ .

5. The sum of 510 and 319 is $510+319 = 829$ . Subtracting -330, we get $829 - (-330) = 829 + 330 = \boxed{1159}$ .

6. Call the unknown number " $\star$ ". Then $\star + 298 = -398$ , so $\star = -398 - 298 = \boxed{-696}$ . (Think about it.)

You might be interested in

Five cupcakes and two cookies cost $19.75. Two cupcakes and four cookies cost $ 17.50.

alina1380 [7]

Answer:

Step-by-step explanation:

Let x and y represent the cost of a cupcake and cookie respectively.

Given that;

Five cupcakes and two cookies cost $19.75.

$5x+2y=19.75 ---------- 1$

Two cupcakes and four cookies cost $17.50.

$2x+4y=17.50 --------------2$

Let's solve the simultaneous equation by elimination;

Multiply equation 1 by 2;

$10x+4y=39.50-------3\\$

Subtract equation 2 from equation 3;

$10x-2x+4y-4y=39.50-8x=22$

$8x=22$

divide both sides by 8

$\frac{8x}{8} =\frac{22}{8} \\x=2.75$

Since we have the value of x, let substitute into equation 1 to get y;

$5x+2y=19.75\\5(2.75)+2y=19.75\\13.75+2y=19.75\\2y=19.75-13.75\\2y=6\\y=\frac{6}{2} \\y=3.00$

therefore , the cost of cupcakes and cookies are;

$cupcakes= 2.75\\cookies= 3.00$

PLEASE MARK ME AS BRAINLIEST

6 0

2 years ago

Just #1 what’s the answer and how do I do it?

Thepotemich [5.8K]

Answer:

It is B

Step-by-step explanation:

(c² - 6c) + (7c + 5)

c² is it's own term

-6c + 7c combine to 'c'

5 is it's own term.

So it is c² + c +5

5 0

2 years ago

Read 2 more answers

Pls help, with explanation if possible

kirza4 [7]

king jon un

Step-by-step explanation:

3 0

2 years ago

Patrick is packing up his closet because he is moving to a new house.He has 100 belts and 25 pairs of shorts to equally distribu

STatiana [176]

It all depends on how big the boxes are but i would say about 5 or 6 boxes if they are big boxes

4 0

3 years ago

Please someone help me to prove this..

Pachacha [2.7K]

Answer: see proof below

<u>Step-by-step explanation:</u>

Use the following Sum to Product Identities:

$\sin x+\sin y=2\sin \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\sin x-\sin y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=2\cos \bigg(\dfrac{x+y}{2}\bigg)\cos \bigg(\dfrac{x-y}{2}\bigg)\\\\\\\cos x+\cos y=-2\sin \bigg(\dfrac{x+y}{2}\bigg)\sin \bigg(\dfrac{x-y}{2}\bigg)$

<u>Proof LHS → RHS</u>

$\text{LHS:}\qquad \qquad \qquad \dfrac{\sin 5-\sin 15+\sin 25 - \sin 35}{\cos 5-\cos 15- \cos 25 + \cos 35}$

$\text{Reqroup:}\qquad \qquad \qquad \dfrac{(\sin 25+\sin 5)-(\sin 35 + \sin 15)}{(\cos 35+\cos 5)-(\cos 25 + \cos 15)}$

$\text{Sum to Product:}\quad \dfrac{2\sin \bigg(\dfrac{25+5}{2}\bigg)\cos \bigg(\dfrac{25-5}{2}\bigg)-2\sin \bigg(\dfrac{35+15}{2}\bigg)\cos \bigg(\dfrac{35-15}{2}\bigg)}{2\cos \bigg(\dfrac{25+15}{2}\bigg)\cos \bigg(\dfrac{25-15}{2}\bigg)-2\cos \bigg(\dfrac{35+5}{2}\bigg)\cos \bigg(\dfrac{35-5}{2}\bigg)}$ $\text{Simplify:}\qquad \qquad \dfrac{2\sin 15\cos 10-2\sin 25\cos 10}{2\cos 20\cos 15-2\cos 20\cos 5}$

$\text{Factor:}\qquad \qquad \dfrac{2\cos 10(\sin 15-\sin 25)}{2\cos 20(\cos 15-\cos 5)}$

$\text{Sum to Product:}\qquad \dfrac{\cos 10\bigg[2\cos \bigg(\dfrac{15+25}{2}\bigg)\sin \bigg(\dfrac{15-25}{2}\bigg)\bigg]}{\cos 20\bigg[-2\sin \bigg(\dfrac{15+5}{2}\bigg)\sin \bigg(\dfrac{15-5}{2}\bigg)\bigg]}$

$\text{Simplify:}\qquad \qquad \dfrac{\cos 10[2\cos 20\sin (-5)]}{\cos 20[-2\sin 10\sin 5]}\\\\\\.\qquad \qquad \qquad =\dfrac{-2\cos10 \cos 20 \sin 5}{-2\sin 10 \cos 20 \sin 5}\\\\\\.\qquad \qquad \qquad =\dfrac{\cos 10}{\sin 10}\\\\\\.\qquad \qquad \qquad =\cot 10$

LHS = RHS: cot 10 = cot 10 $\checkmark$

8 0

2 years ago