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2 years ago

2 1/3 ÷ 1 3/4-2.6÷1.3 pls i rlly need help anything will help

Mathematics

1 answer:

Nat2105 [25]2 years ago

3 0

Answer:

The answer is -2/3

Step-by-step explanation:

Multiply by the reciprocal:

7/3 / 1/4 x 2

Simplify expression:

7/3 x 4/7 -2

Calculate:

1/3 x 4 -2

4/3 - 2

= -2/3

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At Harry's discount hardware everything is sold for 20 percent less than the price marked. If Harry buys tool kits for $80,what

ICE Princess25 [194]

Answer:

$120.

Step-by-step explanation:

The amount he sells the tool kit for = 80 + 20% of 80

= 80 + 16

= $96.

Let m be the marked price, then

m - 0.20m = 96

0.8m = 96

m = $120.

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3 years ago

Harvey collects rare marbles. he has 112 marbles in his collection. each tray holds an equal number of marbles. if he uses 7 tra

WARRIOR [948]

He used 7 trays to hold all the marbles with equal number each tray. That means, there will be 16 marbles for each tray. Therefore, the number of marbles he will display on 3 trays would be a total of 48 marbles. Hope this answers the question. Have a nice day.

3 0

3 years ago

Can someone help me with this? I’ll give brainliest if you answer.

djverab [1.8K]

Answer: yes, they are congruent they are just rotated

Step-by-step explanation:

8 0

3 years ago

What’s the difference for this equation!!!

Katarina [22]

–(x^2–2x+12) would become –x^2+2x–12.

3x^2+9x+13 – x^2+2x–12

= 2x^2 + 11x – 1

7 0

3 years ago

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Determine whether angle ABC should be solved by using the Law of Sines or the Law of Cosines. Then solve the triangle.

drek231 [11]

Answer:

Law of Cosines

Angle A = 45°

Angle B = 51°

Angle C = 84°

Step-by-step explanation:

Law of sines is used when we are given

a) two angles and one side or

b) two sides and non-included angle

Law of cosines is used when we are given

a) three sides or

b) two sides and included angle

In the given question we are given three sides so, Law of Cosines will be used to solve the triangle.

Law of Cosines is:

$a^2 = b^2 + c^2 -2bccos A\\b^2 = a^2 + c^2 -2accos B\\c^2 = a^2 + b^2 -2abcosC$

We will find the three angles A ,B and C of the triangle using above formula.

a= 10, b=11, c=14

Putting values and finding angle A

$a^2 = b^2 + c^2 -2bccos A\\(10)^2 = (11)^2 + (14)^2 -2(11)(14)cosA\\100 = 121 + 196 -308cosA\\100 = 317 -308 cosA\\100-317 = -308cosA\\-217/-308 = cos A\\0.704 = cos A\\=> A = cos ^{-1}(0.704)\\A= 45$

Now finding angle B

$b^2 = a^2 + c^2 -2ac cos B\\(11)^2 = (10)^2 + (14)^2 - 2(10)(14)cosB\\121 = 100+196 - 280cosB\\121 -296 = -280cosB\\-175/-280 = cosB\\0.625 = cosB\\=> B = cos^{-1} (0.625)\\B = 51$

Now finding angle C

$c^2 = a^2 + b^2 -2abcosC\\(14)^2 =(10)^2 + (11)^2 -2(10)(11)cosC\\196 = 100+121 -220cosC\\196 -221 = -220cosC\\-25/-220 = cos C\\0.11 = cosC\\=> C = cos^{-1}(0.11)\\C= 84$

4 0

3 years ago

Read 2 more answers