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3 years ago

Write 6.85• 10 to the 8th power in standard form

Mathematics

2 answers:

zimovet [89]3 years ago

7 0

6.85 x 10^8 = 685,000,000

hope that helps
--------------------------

vagabundo [1.1K]3 years ago

3 0

685,000,000 would be the right number. Hope this helps!

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Expand and combine like terms. (4b^2+ 3) (4b^2- 3) = 1 SEE ANSWER

krok68 [10]

Answer:

Step-by-step explanation:

(4b^2+3)(4b^2-3)

since it is the expanded form of formula a^2-b^2=(a+b)(a-b) we can write,

=(4b^2)^2-(3)^3

=16b^4-9

8 0

2 years ago

2√6 ∙ √8 + 2√12 show your work please and thank you

meriva

Answer:

Step-by-step explanation:

multiply first,

so 2√6 x √8 = 2√48 (because when radicals are multiplied we add them)

2√48 can be simplified to 8√3 and 2√12 can be simplified to 4√3

8√3+4√3=12√3

8 0

3 years ago

Solve for the value of r. (8r+6)° (9r-7)

lukranit [14]

Answer:

r = 13

Step-by-step explanation:

these angles are vertical and are congruent

8r + 6 = 9r - 7

6 = r - 7

13 = r

8 0

2 years ago

Please answer question three and four if you can :) Show full working out ty;)

nikklg [1K]

Step-by-step explanation:

Let D be the mid point of side BC, [B(2, - 1), C(5, 2)].

Therefore, by mid-point formula:

$D = ( \frac{2 + 5}{2}, \: \: \frac{ - 1 + 2}{2} ) = ( \frac{7}{2}, \: \: \frac{ 1}{2} ) \\ \therefore D= (3.5, \: \: 0.5) \\ \& \: A=(-1,\:\:4)...(given) \\\\ now \: by \: distance \: formula \\ \\ Length \: of \: segment \: AD \\ = \sqrt{( - 1 - 3.5)^{2} + {(4 - 0.5)}^{2} } \\ = \sqrt{(4.5)^{2} + {(3.5)}^{2} } \\ = \sqrt{20.25 + 12.25 } \\ = \sqrt{32.5} \\ \red{ \boxed{\therefore Length \: of \: segment \: AD = 5.7 \: units}}$

4 (a)

Equation of line AB[A(2, 1), B(-2, - 11)] in two point form is given as:

$\frac{y-y_1}{y_1-y_2} =\frac{x-x_1}{x_1 - x_2} \\\\\therefore \frac{y-1}{1-(-11)} =\frac{x-2}{2 - (-2) } \\\\\therefore \frac{y-1}{1+11} =\frac{x-2}{2 +2} \\\\\therefore \frac{y-1}{12} =\frac{x-2}{4} \\\\\therefore \frac{y-1}{3} =\frac{x-2}{1} \\\\\therefore y-1= 3(x - 2)\\\\\therefore y= 3x - 6+1\\\\\therefore y= 3x - 5\\\\ \huge \purple {\boxed {\therefore 3x - y-5=0}} \\$

is the equation of line AB.

Now we have to check whether C(4, 7) lie on line AB or not.

Let us substitute x = 4 & y = 7 on the Left hand side of equation of line AB and if it gives us 0, then C lies on the line.

$LHS = 3x - y-5\\=3\times 4-7-5\\= 12-12\\=0\\= RHS$

Hence, point C (4, 7) lie on the straight line AB.

4(b)

Like we did in 4(a), first find the equation of line AB and then substitute the coordinates of point C in equation and if they satisfy the equation, then all the three points lie on the straight line.

4 0

3 years ago

The areas of two similar triangles are 72dm2 and 50dm2. The sum of their perimeters is 226dm. What is the perimeter of each of t

astraxan [27]

Answer:

Hence, the perimeter of the triangles are:

P=123.2727 dm

P'=102.7272 dm

Step-by-step explanation:

In two similar triangles:

The ratio of the areas of two triangle is equal to the square of their perimeters.

Let A and A' represents the area of two triangles and P and P' represents their perimeter.

Then they are related as:

$\dfrac{A}{A'}=\dfrac{P^2}{P'^2}$

We are given:

A=72 dm^2 , A'=50 dm^2

and P+P'=226 dm.-----------(1)

i.e. $\dfrac{72}{50}=\dfrac{P^2}{P'^2}\\\\\dfrac{36}{25}=\dfrac{P^2}{P'^2}$

on taking square root on both the side we get:

$\dfrac{P}{P'}=\dfrac{6}{5}\\\\P=\dfrac{6}{5}P'$

Now putting the value of P in equation (1) we obtain:

$\dfrac{6}{5}P'+P'=226\\\\\dfrac{6P'+5\times P'}{5}=226\\\\\dfrac{6P'+5P'}{5}=226\\\\11P'=226\times 5\\\\11P'=1130\\\\P'=\dfrac{1130}{11}=102.7272$

Hence,

P=226-102.7272=123.2727

Hence, the perimeter of the triangles are:

P=123.2727 dm

P'=102.7272 dm

6 0

3 years ago