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Dovator [93]
3 years ago
10

Without using (10), show directly that (10.6^-1)^8=17.6^-8

Mathematics
1 answer:
blagie [28]3 years ago
3 0
We are given the expression:

(17.6^-1)^8 = 17.6 ^ -8

This means that the two terms on each side are equal. We are asked to show how this is possible. 

First, use the rule of exponents. If a term raised to the power of a number x^n and is further raised to the power m: (x^n)^m, to simplify the expression, multiply n and m and this will be your end exponent = x^nm. 

We can apply this rule here:

17.6^-1 ^ 8

-1 * 8 = -8

then, retain the base and replace the exponent with the product nm:

17.6 ^-8. This proves that the left term is equal to the right term. <span />
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kirza4 [7]

Answer:

331.52 oz

Step-by-step explanation:

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6 0
3 years ago
Read 2 more answers
8.Find the value of c that makes the expression be a perfect square trinomial p^2-30p+c
Darina [25.2K]
Hope this is the right one.

Problem 8
p^2 - 30p + c
<em>Step One</em>
Take 1/2 of - 30
1/2 * -30 = - 15

<em>Step 2
</em>Square -15
(-15)^2 = 225
c = 225

Problem Nine
\text{x = }\dfrac{ -b \pm \sqrt{b^{2} - 4ac } }{2a} &#10;
a = 1
b = 4
c = -15
\text{x = }\dfrac{ -4 \pm \sqrt{4^{2} - 4*1*(-15) } }{2*1}
\text{x = }\dfrac{ -4 \pm \sqrt{\text{16} + \text{60} } }{2}

x = [-4 +/- sqrt(76)] / 2
x = [-4 +/- 2*sqrt19]/2
x = [-4/2 +/- 2/2 sqrt[19]
x = - 2 +/- sqrt(19) 

x1 = - 2 + sqrt(19)
x2 = -2 - sqrt(19)

These two can be broken down more by finding the square root. I will leave them the way they are.  It's just a calculator question if you want it to go into decimal form.

Problem Ten

a = 1
b = 4
c = -32

The discriminate is sqrt(b^2 - 4ac)
D = sqrt(b^2 - 4ac)
D = sqrt(4^2 - 4(1)(-32)
D = sqrt(16 - - 128)
D = sqrt(16 + 128)
D = sqrt(144)
D = +/- 12

Since D can equal + or minus 12 there must be 2 possible (and different) roots. As a matter of fact, this quadratic can be factored.
(x + 8)(x - 4) = y 
But that' s not what you were asked for.
The discriminate is >  0 so the roots are going to be real.
<em>Answer; The discriminate is > 0 so there will be 2 real different roots.</em>

4 0
3 years ago
What is the equation of the following line written in slope intercept form (1,5) (-1,-5)
wlad13 [49]
-5-5/-1-1=
-10/-2=
5

y-5=5(x-1)
y-5=5x-5
y=5x+0
6 0
3 years ago
The lengths of the bases of the right trapezoid are 9 cm and 18 cm. The length of a longer leg is 15 cm. Find the area of the tr
finlep [7]

Firstly, we will draw figure

now, we will draw a altitude from B to DC that divides trapezium into rectangle and right triangle

because of opposite sides of rectangle ABMD are congruent

so,

DM=AB=9

CM=CD-DM

CM=18-9

CM=9

now, we can find BM by using Pythagoras theorem

BM=\sqrt{BC^2-CM^2}

now, we can plug values

we get

BM=\sqrt{15^2-9^2}

BM=12

now, we can find area of trapezium

A=\frac{1}{2} (AB+CD)*(BM)

now, we can plug values

and we get

A=\frac{1}{2} (9+18)*(12)

A=162cm^2

so, area of of the trapezoid is 162cm^2..........Answer

6 0
3 years ago
How do you simplify expressions with rational exponents
Rainbow [258]

Answer:

Step-by-step explanation:

Simplify expression with rational exponents can look like a huge thing when you first see them with those fractions sitting up there in the exponent but let's remember our properties for dealing with exponents. We can apply those with fractions as well.

Examples

(a)   (p^4)^{\dfrac{3}{2}}

From above, we have a power to a power, so, we can think of multiplying the exponents.

i.e.

(p^{^ {\dfrac{4}{1}}})^{\dfrac{3}{2}}

(p^{^ {\dfrac{12}{2}}})

Let's recall that when we are dealing with exponents that are fractions, we can simplify them just like normal fractions.

SO;

(p^{^ {\dfrac{12}{2}}})

= (p^{ 6})

Let's take a look at another example

\Bigg (27x^{^\Big{6}} \Bigg) ^{{\dfrac{5}{3}}}

Here, we apply the \dfrac{5}{3} to both 27 and x^6

= \Bigg (27^{{\dfrac{5}{3}}} \times x^\Big{\dfrac{6}{1}\times {{\dfrac{5}{3}}} }\Bigg)

= \Bigg (27^{{\dfrac{5}{3}}} \times x^\Big{\dfrac{2}{1}\times {{\dfrac{5}{1}}} }\Bigg)

Let us recall that in the rational exponent, the denominator is the root and the numerator is the exponent of such a particular number.

∴

= \Bigg (\sqrt[3]{27}^{5} \times x^{10} }\Bigg)

= \Bigg (3^{5} \times x^{10} }\Bigg)

= 249x^{10}

8 0
2 years ago
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