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3 years ago

9

I have 7 hundreds blocks, 5 tens blocks, and 8 ones blocks. I use my blocks to model two 3-digit numbers. What could my two numb

er be?

1 answer:

never [62]3 years ago

6 0

Answer:

758

700+50+8=758

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What number can go into 2470,121 and 45<br>

Usimov [2.4K]

Answer:

Step-by-step explanation:

All of these numbers can be divided by 2, 3, 5, 7, 11, 13...

There is no whole number if you want to divide each number.

121 only goes into 1, 11, and 121. No other numbers have 11

45/2 = 22.5

45/3 = 15

45/5 = 9

2470/2 = 1235

2470/3 = 823.33333...

2470/5 = 494

121/2 = 60.5

121/3 = 40.33333...

121/5 = 24.2

4 0

3 years ago

Where A = -7 and b = 9 what is the value of the midpoint?

sergey [27]

Answer:

Find the value of x if B is the midpoint of AC, AB = 2x + 9 and BC = 37

Step-by-step explanation:

8 0

3 years ago

Find the equation of the line.<br><br>Use exact numbers.

Taya2010 [7]

Slope intercept form: y = mx + b

m = slope

b = y-intercept

The line passes through the y-axis at the point (0, -5). So, -5 is the y-intercept.

We can use the points (0, -5) and (5, 0) to solve.

Slope formula: y2-y2/x2-x1

0-(-5)/5-0)

5/5

1

The slope is 1.

Substitute these values into the equation.

y = 1x + (-5)

y = x - 5

Therefore, the answer is [ y = x - 5 ]

Best of Luck!

6 0

3 years ago

A sign on the pumps at a gas station encourages customers to have their oil checked, and claims that one out of 10 cars needs to

Archy [21]

Answer:

0.2916, 0.1488, 0.0319

Step-by-step explanation:

Given that a sign on the pumps at a gas station encourages customers to have their oil checked, and claims that one out of 10 cars needs to have oil added.

Since each trial is independent there is a constant probability for any random car to need oil is 0.10

Let X be the number of cars that need oil

A) Here X is BIN(4,0.1)

$P(X=1) = 4C1(0.1)(0.9)^3 \\= 0.2916$

B) Here X is Bin (8, 0.1)

$P(x=2) = 8C2 (0.1)^2 (0.9)^6\\\\=0.1488$

C) Here X is Bin (20,5)

$P(x=5) = 20C5 (0.1)^5 (0.9)^{15} \\=0.0319$

3 0

3 years ago

The longer leg of a right triangle is 6cm longer than the shorter leg and also 1cm longer than the hyptenese. find the perimeter

CaHeK987 [17]

X - a leg (x > 6);
x - 6 - a leg;
x + 1 - a <span>hypotenuse;</span>
Use the Pythagorean theorem
$(x+1)^2=x^2+(x-6)^2$

Use: $(a\pm b)^2=a^2\pm2ab+b^2$

$x^2+2x+1=x^2+x^2-12x+36\ \ \ |-x^2-2x-1\\\\x^2-14x+35=0\ \ \ |-35\\\\x^2-14x=-35\\\\x^2-2\cdot x\cdot7=-35\ \ \ |+7^2\\\\x^2-2\cdot x\cdot7+7^2=-35+7^2\\\\(x-7)^2=14\iff x-7=\pm\sqrt{14}\ \ \ |+7\\\\x=7-\sqrt{14} < 6;\ x=7+\sqrt{14}$

$x=7+\sqrt{14}\\x-6=1+\sqrt{14}\\x+1=8+\sqrt{14}$

The perimeter:

$7+\sqrt{14}+1+\sqrt{14}+8+\sqrt{14}=16+3\sqrt{14}$

5 0

3 years ago