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2 years ago

7

The Martinez family has $1543 a month income. They have $978 and fixed expenses, money they have to spend. How much money do the

y have left after the fixed expenses

1 answer:

KengaRu [80]2 years ago

8 0

Answer:

$565

Step-by-step explanation:

1543-978=565

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Help please jggvnhujff

saveliy_v [14]

Answer:

Step-by-step explanation:

<u><em>8).</em></u> $\left \{ {{4x+3y=1} \atop {x+y=2}} \right.$

<em>(2)</em> × [ - 3 ]

4x + 3y = 1 ........ <em>(3)</em>

- 3x - 3y = - 6 .... <em>(4)</em>

<em>(3)</em> + <em>(4)</em>

x = - 5

- 5 + y = 2 ⇒ y = 7

<em>( - 5 , 7 )</em>

<u><em>9).</em></u> $\left \{ {{-9x+3y=18} \atop {2x+y=-4}} \right.$

<em>(1)</em> ÷ [- 3]

3x - y = - 6 ......... <em>(3)</em>

2x + y = - 4 ........ <em>(4)</em>

<em>(3)</em> + <em>(4)</em>

5x = - 10 ⇒ x = - 2

2(- 2) + y = - 4 ⇒ y = 0

<em>(- 2, 0)</em>

<u><em>10).</em></u> $\left \{ {{x-2y=14} \atop {10x-6y=0}} \right.$

<em>(2)</em> ÷ 10

x - 0.6y = 0 ⇒ x = 0.6y -----> <em>(1)</em>

0.6y - 2y = 14

- 1.4y = 14

y = - 10

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<em>(- 6, - 10)</em>

Now is your turn, you can do it!!

3 0

3 years ago

A technical machinist is asked to build a cubical steel tank that will hold of water. Calculate in meters the smallest possible

Charra [1.4K]

Answer:

The smallest possible length is 0.83m

Step-by-step explanation:

Given

$Volume = 565L$

Required

The smallest length of the tank

Since the tank is cubical, then the volume is:

$Volume = Length^3$

This gives:

$565L= Length^3$

Express as $m^3$

$\frac{565m^3}{1000} = Length^3$

$0.565m^3 = Length^3$

Take cube roots of both sides

$0.8267m = Length$

Rewrite as:

$Length = 0.8267m$

Approximate

$Length = 0.83m$

5 0

3 years ago

Suppose we have 3 cards identical in form except that both sides of the first card are colored red, both sides of the second car

Nikolay [14]

Answer:

probability that the other side is colored black if the upper side of the chosen card is colored red = 1/3

Step-by-step explanation:

First of all;

Let B1 be the event that the card with two red sides is selected

Let B2 be the event that the

card with two black sides is selected

Let B3 be the event that the card with one red side and one black side is

selected

Let A be the event that the upper side of the selected card (when put down on the ground)

is red.

Now, from the question;

P(B3) = ⅓

P(A|B3) = ½

P(B1) = ⅓

P(A|B1) = 1

P(B2) = ⅓

P(A|B2)) = 0

(P(B3) = ⅓

P(A|B3) = ½

Now, we want to find the probability that the other side is colored black if the upper side of the chosen card is colored red. This probability is; P(B3|A). Thus, from the Bayes’ formula, it follows that;

P(B3|A) = [P(B3)•P(A|B3)]/[(P(B1)•P(A|B1)) + (P(B2)•P(A|B2)) + (P(B3)•P(A|B3))]

Thus;

P(B3|A) = [⅓×½]/[(⅓×1) + (⅓•0) + (⅓×½)]

P(B3|A) = (1/6)/(⅓ + 0 + 1/6)

P(B3|A) = (1/6)/(1/2)

P(B3|A) = 1/3

5 0

3 years ago

1) Simplify the following expressions: a) sin²108° + sin² 18°

abruzzese [7]

Note that

108° = 90° + 18°

so

sin(108°) = sin(90° + 18°) = sin(90°) cos(18°) + cos(90°) sin(18°) = cos(18°)

Then

sin²(108°) + sin²(18°) = cos²(18°) + sin²(18°) = 1

by the Pythagorean identity.

8 0

2 years ago

Create a memory device to help you remember the quadratic formula. The device could be a song, drawing, rhyme, mnemonic device,

Lostsunrise [7]

Answer:

q-u-a-d-r-a-t-i-c f-o-r-m-u-l-a

Step-by-step explanation:

6 0

3 years ago