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4 years ago

It’s solve system by elimination!!! plz help asap

Mathematics

1 answer:

zhuklara [117]4 years ago

3 0

Answer:

look at the picture shown

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brine's kitchen has an area of 81 square feet. the kitchen is 9 times as many square feet as brine's pantry. if the pantry is 3

maks197457 [2]

So 81/9=9 so area of pantry is 9

3 times x=9
3x=9
divide both sides by 3
x=3

legnth/width=3

8 0

3 years ago

An umbrella is made by stitching 10 triangular pieces of clothes of 2 different colors, each piece measuring 9 cm, 40 cm and 41

malfutka [58]

Answer:

See below in bold.

Step-by-step explanation:

The triangular piece of cloth is a right triangle because 41^2 = 40^2 + 9^2 (By the Pythagoras theorem).

So the area of one piece = 1/2 * 9 * 40 cm^2.

The total amount of cloth required for one umbrella = 10 * 1/2 * 9 * 40

= 1800 cm^2.

The total cost = 1800 * 0.25 = Rs 450.

3 0

3 years ago

A student's grade depends on her score on four equally-weighted exams. Her average on the first three exams is 92. What must she

Sergeeva-Olga [200]

3x92 = 276 (first 3 exams)
average of at least 90 (90 x 4 = 360)

360 - 276 = 84

Answer
she needs to score at least 84 on the fourth exam in order to guarantee a final average of at least 90

4 0

3 years ago

Add. (6x3+3x2−2)+(x3−5x2−3) Express the answer in standard form.

sergiy2304 [10]

Answer:

$7x^3-2x^2-5$

Step-by-step explanation:

We need to add the two terms.

$(6x^3+3x^2-2)+(x^3-5x^2-3)$

Solving,

Combine the like terms and adding those terms

$(6x^3+3x^2-2)+(x^3-5x^2-3)\\=6x^3+3x^2-2+x^3-5x^2-3\\=6x^3+x^3+3x^2-5x^2-2-3\\=7x^3-2x^2-5$

So, the answer is:

$7x^3-2x^2-5$

6 0

4 years ago

Please someone what is the total surface area of this pryamid

Dimas [21]

Answer:

$\large\boxed{C.\ (25+25\sqrt3)\ in^2}$

Step-by-step explanation:

We have the square and four equilateral triangles.

The formula of an area of a squre:

$A_S=a^2$

a - length of side

The formula of an area of an equilateral triangle:

$A_T=\dfrac{a^2\sqrt3}{4}$

a - length of side

Clculate the areas:

SQURE:

$A_S=x^2\ in^2$

TRIANGLE:

$A_T=\dfrac{x^2\sqrt3}{4}\ in^2$

The SURFACE AREA of a square pyramid:

$S.A.=A_S+4A_T\\\\S.A.=x^2+4\cdot\dfrac{x^2\sqrt3}{4}=x^2+x^2\sqrt3$

Put x = 5:

$S.A.=5^2+5^2\sqrt3=(25+25\sqrt3)\ in^2$

7 0

3 years ago