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3 years ago

Find the length of AB if AC=18 and BC=7

Mathematics

1 answer:

kotegsom [21]3 years ago

3 0

Answer:

Step-by-step explanation:

Step 1:

AC + BC = AB Equation

Step 2:

18 + 7 = AB Substitute

Answer:

25 Add

Hope This Helps :)

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The product of 1 1/2 and 2 is

Over [174]

Method 1:

Convert the mixed number to the improper fraction:

$1\dfrac{1}{2}=\dfrac{1\cdot2+1}{2}=\dfrac{3}{2}$

Make the product:

$1\dfrac{1}{2}\cdot2=\dfrac{3}{2}\cdot2$

<em>canceled 2</em>

$=\dfrac{3}{\not2_1}\cdot\not2^1}=\boxed{3}$

Method 2:

$1\dfrac{1}{2}=1+\dfrac{1}{2}$

$1\dfrac{1}{2}\cdot2=\left(1+\dfrac{1}{2}\right)\cdot2$

<em>use the distributive property a(b + c) = ab + ac</em>

$(1)(2)+\left(\dfrac{1}{2}\right)(2)=2+1=\boxed{3}$

7 0

3 years ago

Write a 3-digit number that is divisible by both 3 and 4. Explain how you know this number is divisible by 3 and 4.

Alexeev081 [22]

Answer:

let us assume the 3 digit number be 984

To check weather the number is divisible by 3 and 4 then when should know the divisibility rule of 3 and 4.

A number is said to be divisible by 3 when the sum of each digit of the number is divisible by 3. Example 9+8+4 = 21 is divisible by 3 hence the number is divisible by 3.

A number is said to be divisible by 4 when the last two digit is divisible by 4 like 84 is divisible by 4 hence 984 is divisible by 4.

5 0

3 years ago

Read 2 more answers

Find all points on the portion of the plane x+y+z=5 in the first octant at which f(x, y, z) = xy2z2 has a maximum value.

irina1246 [14]

Lagrange multipliers:

$L(x,y,z,\lambda)=xy^2z^2+\lambda(x+y+z-5)$

$L_x=y^2z^2+\lambda=0$
$L_y=2xyz^2+\lambda=0$
$L_z=2xy^2z+\lambda=0$
$L_\lambda=x+y+z-5=0$

$\lambda=-y^2z^2=-2xyz^2=-2xy^2z$

$-y^2z^2=-2xyz^2\implies y=2x$ (if $y,z\neq0$ )

$-y^2z^2=-2xy^2z\implies z=2x$ (if $y,z\neq0$ )

$-2xyz^2=-2xy^2z\implies z=y$ (if $x,y,z\neq0$ )

In the first octant, we assume $x,y,z>0$ , so we can ignore the caveats above. Now,

$x+y+z=5\iff x+2x+2x=5x=5\implies x=1\implies y=z=2$

so that the only critical point in the region of interest is (1, 2, 2), for which we get a maximum value of $f(1,2,2)=16$ .

We also need to check the boundary of the region, i.e. the intersection of $x+y+z=5$ with the three coordinate axes. But in each case, we would end up setting at least one of the variables to 0, which would force $f(x,y,z)=0$ , so the point we found is the only extremum.