Can someone please help me with 6-12 please and i’m being timed

How can 7 1/5+(-8 3/5) be expressed as the sum of its integer and fractional parts

Nadya [2.5K]

Answer:

see the explanation

Step-by-step explanation:

we know that

A mixed number is the sum of a integer and a fractional part

we have

$7\frac{1}{5}+(-8\frac{3}{5})$

so

$7\frac{1}{5}=7+\frac{1}{5}$

$-8\frac{3}{5}=-8-\frac{3}{5}$

substitute

$7+\frac{1}{5}+(-8-\frac{3}{5})$

Group terms

$(7-8)+(\frac{1}{5}-\frac{3}{5})$

$(-1)+(-\frac{2}{5})$

$-1\frac{2}{5}$

6 0

3 years ago

Marisols bedroom has an area of 29. 76 square meters. the length of the room is 6. 2 meters. what’s is the width

andriy [413]

Answer:

4.8 m

Step-by-step explanation:

The area of a rectangle is the product of length and width. This relation can be used to find any of the dimensions when the other two are known.

__

Here, we know area and length, so we can find width.

A = LW

W = A/L = (29.76 m²)/(6.2 m) = 4.8 m

Mariosol's bedroom is 4.8 meters wide.

7 0

3 years ago

Two dice are rolled; find the probability for total: a) 1, b) 4 or 6; c) <13

IrinaVladis [17]

Answer:

0 ; 2/9 ; 1

Step-by-step explanation:

In a roll of two dice ;

The sample space = (Number of faces on die)^number of dice rolled = 6^2 = 36

The total possible outcome = sample space = 36

The probability of an event (A) :

P(A) = number of required outcome / number of total possible outcomes

find the probability for total: a) 1,

P(total of 1)

Number of 1 total = 0

Hence, P(total of 1) = 0

b) 4 or 6;

Number of 4 total = 3

Number of 6 total = 5

3 /36 + 5 / 36 = 8/36 = 2/9

c) <13

P(total < 13)

Number of total < 13 = 36

36 / 36 = 1

3 0

3 years ago

Find the valueof the symbol x: (x÷5)×4=80÷(5×4)

Strike441 [17]

Answer:

Step-by-step explanation:

(x÷5)*4=80÷(5*4)

(x÷5)*4 = 80÷20

(x÷5)*4 = 4

x÷5 = 4÷4

$\frac{x}{5}=1\\\\x=1*5\\\\x=5$

4 0

3 years ago

Read 2 more answers

Evaluate integral _C x ds, where C is

borishaifa [10]

Answer:

a. $\mathbf{36 \sqrt{5}}$

b. $\mathbf{ \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}$

Step-by-step explanation:

Evaluate integral _C x ds where C is

a. the straight line segment x = t, y = t/2, from (0, 0) to (12, 6)

i . e

$\int \limits _c \ x \ ds$

where;

x = t , y = t/2

the derivative of x with respect to t is:

$\dfrac{dx}{dt}= 1$

the derivative of y with respect to t is:

$\dfrac{dy}{dt}= \dfrac{1}{2}$

and t varies from 0 to 12.

we all know that:

$ds=\sqrt{ (\dfrac{dx}{dt})^2 + ( \dfrac{dy}{dt} )^2}} \ \ dt$

∴

$\int \limits _c \ x \ ds = \int \limits ^{12}_{t=0} \ t \ \sqrt{1+(\dfrac{1}{2})^2} \ dt$

$= \int \limits ^{12}_{0} \ \dfrac{\sqrt{5}}{2}(\dfrac{t^2}{2}) \ dt$

$= \dfrac{\sqrt{5}}{2} \ \ [\dfrac{t^2}{2}]^{12}_0$

$= \dfrac{\sqrt{5}}{4}\times 144$

= $\mathbf{36 \sqrt{5}}$

b. the parabolic curve x = t, y = 3t^2, from (0, 0) to (2, 12)

Given that:

x = t ; y = 3t²

the derivative of x with respect to t is:

$\dfrac{dx}{dt}= 1$

the derivative of y with respect to t is:

$\dfrac{dy}{dt} = 6t$

$ds = \sqrt{1+36 \ t^2} \ dt$

Hence; the integral _C x ds is:

$\int \limits _c \ x \ ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \ dt$

Let consider u to be equal to 1 + 36t²

1 + 36t² = u

Then, the differential of t with respect to u is :

76 tdt = du

$tdt = \dfrac{du}{76}$

The upper limit of the integral is = 1 + 36× 2² = 1 + 36×4= 145

Thus;

$\int \limits _c \ x \ ds = \int \limits _0 \ t \ \sqrt{1+36 \ t^2} \ dt$

$\mathtt{= \int \limits ^{145}_{0} \sqrt{u} \ \dfrac{1}{72} \ du}$

$= \dfrac{1}{72} \times \dfrac{2}{3} \begin {pmatrix} u^{3/2} \end {pmatrix} ^{145}_{1}$

$\mathtt{= \dfrac{2}{216} [ 145 \sqrt{145} - 1]}$

$\mathbf{= \dfrac{1}{108} [ 145 \sqrt{145} - 1]}}$

5 0

4 years ago