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3 years ago

Which are partial products for 68 × 43?

Mathematics

1 answer:

IrinaVladis [17]3 years ago

6 0

The product 68 × 43 can be written as the sum ...

... 68 × 43 = 60×40 + 60×3 + 8×40 + 8×3

Of these four partial products, only two are listed in your list of answer choices:

... B. 8 × 3 = 24

... C. 8 × 40 = 320

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Add. Write fractions in simplest form.<br> 4-12 EVENS

Virty [35]

4. 1/3 6.-0.9 or -9/10 8. 2 1/3 10. -3.45 12.4.54 or 4 27/50

7 0

3 years ago

A pet store sells goldfish and hermit crabs,

Elis [28]

Answer:

$Cost = \$28$

Step-by-step explanation:

Given

Represent Goldfish with g and hermit crabs with h.

The first statement, we have:

$7g + 3h = 26$

The second statement, we have:

$4g + 5h = 28$

Required

Determine the selling price of 6 goldfish and 4 hermit crabs

The equations are:

$7g + 3h = 26$ --- (1)

$4g + 5h = 28$ --- (2)

Make g the subject in (2)

$4g + 5h = 28$

$4g = 28 - 5h$

Divide both sides by 4

$g = \frac{1}{4}(28 - 5h)$

Substitute $\frac{1}{4}(28 - 5h)$ for g in (1)

$7g + 3h = 26$

$7(\frac{1}{4}(28 - 5h)) + 3h = 26$

$\frac{7}{4}(28 - 5h) + 3h = 26$

Multiply through by 4

$4 * \frac{7}{4}(28 - 5h) + 4*3h = 26*4$

$7(28 - 5h) + 4*3h = 26*4$

Open bracket

$196 - 35h + 12h = 104$

$196 -23h = 104$

Collect Like Terms

$-23h = 104-196$

$-23h = -92$

Make h the subject

$h = \frac{-92}{-23}$

$h = \frac{92}{23}$

$h = 4$

Substitute 4 for h in $g = \frac{1}{4}(28 - 5h)$

$g = \frac{1}{4}(28 - 5*4)$

$g = \frac{1}{4}(28 - 20)$

$g = \frac{1}{4}(8)$

$g = 2$

This implies that:

1 goldfish = $2

1 hermit crab = $4

The cost of 6 goldfish and 4 hermit crabs is:

$Cost = 6g + 4h$

$Cost = 6*\$2 + 4*\$4$

$Cost = \$12 + \$16$

$Cost = \$28$

5 0

3 years ago

Leno4ka [110]

Answer:

$\frac{s^2-25}{(s^2+25)^2}$

Step-by-step explanation:

Let's use the definition of the Laplace transform and the identity given: $\mathcal{L}[t \cos 5t]=(-1)F'(s)$ with $F(s)=\mathcal{L}[\cos 5t]$ .

Now, $F(s)=\int_0 ^{+ \infty}e^{-st}\cos(5t) dt$ . Using integration by parts with u=e^(-st) and dv=cos(5t), we obtain that $F(s)=\frac{1}{5}\sin(5t)e^{-st} |_{0}^{+\infty}+\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt=\int_0 ^{+ \infty}e^{-st}\sin(5t) dt$ .

Using integration by parts again with u=e^(-st) and dv=sin(5t), we obtain that

$F(s)=\frac{s}{5}(\frac{-1}{5}\cos(5t)e^{-st} |_{0}^{+\infty}-\frac{s}{5}\int_0 ^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}(\frac{1}{5}-\frac{s}{5}\int_0^{+ \infty}e^{-st}\sin(5t) dt)=\frac{s}{5}-\frac{s^2}{25}F(s)$ .