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3 years ago

___m = 1km _ml = 1DL ___cg = 1dag

1 answer:

6 0

( 1,000m = 1km )
—There is 1 KILOMETER is 1000 meters.

(100ml = 1DL)
<span>—There are 100 milliliters in one deciliter, and there are 10 milliliters in 0.1 deciliters.
</span>
( 1,000cg = 1dag )
—There are 1,000 centigrams in 1 decagram, and there are 0.001 decagrams in one centigram.

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Find the length of the missing side and leave the answer in simplest radical form.

Alex73 [517]

Answer:

3√5 ft

Step-by-step explanation:

third side=√(9²-6²)=√(81-36)=√(45)=3√5 ft

8 0

3 years ago

Read 2 more answers

Jerry is paving the parking spot shown below. It costs \$50$50dollar sign, 50 to pave each square meter.

EastWind [94]

Answer:

$1250

Step-by-step explanation:

6 1/4 × 4 = 25

25 × 50 = cost

Cost = $1250

5 0

4 years ago

Read 2 more answers

Suppose your firm had the following taxable income amounts: 2015 ($5 million) operating loss 2016 $4 million 2017 $4 million 201

ioda

Answer:

$7,000,000

Step-by-step explanation:

The taxable income is the amount of money (income earned or unearned) by an individual or an organization that creates a potential tax liability.

The formula is shown below:

Taxable Income Formula = Gross Total Income – Total Exemptions – Total Deductions

Note that $5,000,000 for 2015 is an operating loss which is a deduction.

Gross Total Income = $4000000+$4000000+$4000000 = $12,000,000

Total Exemption = 0

Total Deductions = $5,000,000

Taxable Income Formula = $12,000,000 - $5,000,000 = $7,000,000

5 0

3 years ago

(a)5%)Let fx, y) = x^4 + y^4 - 4xy + 1. and classify each critical point Find all critical points of fx,y) as a local minimum, l

lara31 [8.8K]

$f(x,y)=x^4+y^4-4xy+1$

has critical points wherever the partial derivatives vanish:

$f_x=4x^3-4y=0\implies x^3=y$

$f_y=4y^3-4x=0\implies y^3=x$

Then

$x^3=y\implies x^9=x\implies x(x^8-1)=0\implies x=0\text{ or }x=\pm1$

If $x=0$ , then $y=0$ ; critical point at (0, 0)
If $x=1$ , then $y=1$ ; critical point at (1, 1)
If $x=-1$ , then $y=-1$ ; critical point at (-1, -1)

$f(x,y)$ has Hessian matrix

$H(x,y)=\begin{bmatrix}12x^2&-4\\-4&12y^2\end{bmatrix}$

with determinant

$\det H(x,y)=144x^2y^2-16$

At (0, 0), the Hessian determinant is -16, which indicates a saddle point.
At (1, 1), the determinant is 128, and $f_{xx}(1,1)=12$ , which indicates a local minimum.
At (-1, -1), the determinant is again 128, and $f_{xx}(-1,-1)=12$ , which indicates another local minimum.