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Katyanochek1 [597]

3 years ago

9

Use base ten blocks to find 182 ÷14 describe the steps you took to find your answer

1 answer:

docker41 [41]3 years ago

4 0

It would be
one 10 block
and 3 single blocks

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Select all the statements that describe the expression 5 + 2x.

FromTheMoon [43]

It's A that's correct!!!!

8 0

3 years ago

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What's the awnswer to 10 divided by 2/7?

Dvinal [7]

34.48

(2/7=.29, 10/.29=34.48)

4 0

3 years ago

Which two values of x are roots of the polynomial below?<br> 3x2-3x+1

QveST [7]

Answer:

The roots are

$x=\frac{1}{6} [3+ i\sqrt{3}]$

$x=\frac{1}{6} [3- i\sqrt{3}]$

Step-by-step explanation:

we have

$3x^2-3x+1$

The formula to solve a quadratic equation of the form

$ax^{2} +bx+c=0$

is equal to

$x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}$

in this problem we have

$3x^2-3x+1=0$

so

$a=3\\b=-3\\c=1$

substitute in the formula

$x=\frac{-(-3)\pm\sqrt{-3^{2}-4(3)(1)}} {2(3)}$

$x=\frac{3\pm\sqrt{-3}} {6}$

Remember that

$i=\sqrt{-1}$

so

$x=\frac{1}{6} [3\pm i\sqrt{3}]$

The roots are

$x=\frac{1}{6} [3+ i\sqrt{3}]$

$x=\frac{1}{6} [3- i\sqrt{3}]$

6 0

3 years ago

Y−3=−1/2(x+4) I need help

frozen [14]

Answer: The slope is - 1/2.

Step-by-step explanation:

4 0

3 years ago

1.- What is 'a' for this hyperbola?

SpyIntel [72]

Answer:

The standard form of a hyperbola with vertices and foci on the x-axis:

$\dfrac{(x-h)^2}{a^2}-\dfrac{(y-k)^2}{b^2}=1$

where:

center: (h, k)
vertices: (h+a, k) and (h-a,k)
Foci: (h+c, k) and (h-c, k) where the value of c is c² = a² + b²
Slopes of asymptotes: $\pm\left(\dfrac{b}{a}\right)$

<h3><u>Part 1</u></h3>

The center of the given hyperbola is (0, 0), therefore:

$\implies \dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$

Therefore $(\pm a,0)$ are the vertices. From inspection of the graph, $a=2$ .

<h3><u>Part 2</u></h3>

Choose two points on the asymptote with the positive slope:

(0, 0) and (4, 6)

Use the slope formula to find the slope:

$\sf slope\:(m)=\dfrac{change\:in\:y}{change\:in\:x}=\dfrac{6-0}{4-0}=\dfrac{3}{2}$

<h3><u>Part 3</u></h3>

Use the <u>slopes of asymptotes</u> formula, compare with the slope found in part 2:

$\implies \dfrac{b}{a}=\dfrac{3}{2}$

Therefore, $b=3$

<h3><u>Part 4</u></h3>

Substitute the found values of $a$ and $b$ into the equation from part 1:

$\implies \dfrac{x^2}{2^2}-\dfrac{y^2}{3^2}=1$

$\implies \dfrac{x^2}{4}-\dfrac{y^2}{9}=1$

4 0

1 year ago