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Mathematics

1 answer:

Sever21 [200]3 years ago

6 0

Answer:

The approximate area of the cross section passing through the diagonals of opposite faces of this cube is:

C. <u>35 square centimeters</u>.

Step-by-step explanation:

When you think in the cross section passing through the diagonals of opposite faces of the cube given, you can see that it's a rectangle with long = 7 centimeters and wide = 5 centimeters ( the measure of an edge of the cube), having this, you can calculate the area:

Area of a rectangle = long * wide
Area of a rectangle = 7 cm * 5 cm
<u>Area of a rectangle = 35 cm^2 or 35 square centimeters</u>

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Which expression is equivalent to t+4+3-2

FrozenT [24]

Answer:

t+5

Step-by-step explanation:

We can combine like terms, and get t+(4+3-2) to get t+5.

Hope this helped!

3 0

3 years ago

Read 2 more answers

I'd really appreciate some help on these questions, I'm really struggling

nikklg [1K]

We can use linear combinations of the equations to eliminate variables.

3x - 4y = 1
-2x + 3y = 1

To eliminate y we'll make the linear combination of 3 times the first equation minus four times the second.

9x - 12y = 3
-8x + 12y = 4

Adding,
x = 7

We could solve for y directly but let's use another linear combination, twice the first plus three times the second:

2(3x - 4y) + 3(-2x + 3y)= 2(1)+3(1)

y = 5

Check: 3(7)-4(5)=1 good. -2(7)+3(5)=1 good.

Q18 Answer: (7,5)

y = -3x + 5
5x - 4y = -3

4y +1(5x - 4y) = 4(-3x + 5) + 1(-3)

5x = -12x + 20 -3

17 x = 17

x = 1
y = -3(1) + 5 = 2
Check: 5(1) - 4(2) = -3 good

Q19 Answer (1,2)

6x + 5y = 25
x = 2y + 24
6x = 12y + 144
5y = 25 - 12y - 144
17y = -119
y = -119/17= -7
x = 2y+24= 10
Check: 6(10)+5(-7)=25 good 2y+24=2(-7)+24=10=x good

Q20 Answer (10,-7)

3x + y = 18
-7x + 3y = -10
9x + 3y = 54
9x - -7x = 54 - -10
16x = 64
x=4
y = 18 -3x = 18-12=6

Check: 3(4)+6=18 good, -7(4)+3(6)=-10 good

Q21 Answer: (4,6)

6 0

3 years ago

The 10th term of an a.p is -27 and the 5th term is -12 .find the first term?

horsena [70]

Step-by-step explanation:

$a_{10} = - 27... (Given) \\\\\therefore a + 9d = - 27...... (1)\\\\a_{5} = - 12... (Given) \\\\\therefore a + 4d = - 12.......(2)\\\\$

Subtracting equation (2) from equation (1)

$a + 9d-( a + 4d )= - 27-(-12)\\\\\therefore a +9d - a-4d = - 27+12\\\\\therefore 5d = - 15\\\\\therefore d = \frac {- 15}{5}\\\\\huge \orange {\boxed {\therefore d = -3}} \\\\$