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3 years ago

Y=4.5 when x=3 find the variation of k

Mathematics

1 answer:

torisob [31]3 years ago

5 0

The variation of K is what is found in front of X

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10(2^×) + 7(3^×) = 6^× + 70

Mkey [24]

Answer:

x = 2.81 and 2.096

Step-by-step explanation:

Given the expression

10(2^x) + 7(3^x) = 6^x + 70

This can also be expressed as;

10(2^x) + 7(3^x) = (2*3)^x + 70

10(2^x) + 7(3^x) = 2^x*3^x + 70

Let a = 2^x and b = 3^x

10a + 7b = ab + 70

10a + 7b - ab = 70

10a-ab + 7b - 70 = 0

a(10-b)+7(b-10) = 0

a(10-b)-7(10-b) = 0

a-7 = 0 and 10-b = 0

a = 7 and b = 10

Since a = 2^x

7 = 2^x

log 7 = log2^x

log7 = xlog2

x = log7/log2

x = 2.81

Similarly

10 = 3^x

log 10 = log 3^x

log 10 = xlog3

x = log 10/log 3

x = 1/0.4771

x = 2.096

Hence the values of x that satisfies the equation are 2.81 and 2.096

8 0

2 years ago

Lillian went shopping for a new phone because of a sale. The price on the tag was $20, but Lillian paid $16 before tax. Find the

ycow [4]

Answer:

20 % discount

Step-by-step explanation:

Lilian got 4 dollars off of the original price (20 - 16 = 4 )

4 / 20 * 100% = 20% off

3 0

2 years ago

There are about 795 house in this region . the average family size is 6 people . there

jekas [21]

Answer:

4770 people

Step-by-step explanation:

795*6= 4770

7 0

3 years ago

I need help please???

alexdok [17]

Answer:

x<29

Step-by-step explanation:

This would be the BEST answer. Hope it helps! ☺

4 0

2 years ago

Write an equation that has a graph with the shape of y = x^2, but shifted left 3 units.

svetoff [14.1K]

$\bf \qquad \qquad \qquad \qquad \textit{function transformations}
\\ \quad \\\\
% left side templates
\begin{array}{llll}
f(x)=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
y=&{{ A}}({{ B}}x+{{ C}})+{{ D}}
\\ \quad \\
f(x)=&{{ A}}\sqrt{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}}(\mathbb{R})^{{{ B}}x+{{ C}}}+{{ D}}
\\ \quad \\
f(x)=&{{ A}} sin\left({{ B }}x+{{ C}} \right)+{{ D}}
\end{array}\\\\
--------------------\\\\$

$\bf % template detailing
\bullet \textit{ stretches or shrinks horizontally by } {{ A}}\cdot {{ B}}\\\\
\bullet \textit{ flips it upside-down if }{{ A}}\textit{ is negative}
\\\\
\bullet \textit{ horizontal shift by }\frac{{{ C}}}{{{ B}}}\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is negative, to the right}\\\\
\left. \qquad \right. if\ \frac{{{ C}}}{{{ B}}}\textit{ is positive, to the left}\\\\$

$\bf \bullet \textit{ vertical shift by }{{ D}}\\
\left. \qquad \right. if\ {{ D}}\textit{ is negative, downwards}\\\\
\left. \qquad \right. if\ {{ D}}\textit{ is positive, upwards}\\\\
\bullet \textit{ period of }\frac{2\pi }{{{ B}}}$

now, with that template in mind, let's see

$\bf y=x^2\implies 
\begin{array}{llll}
y=(&1x&-0)^2\\
&B&C
\end{array}$

so, just change C to +3, thus C/B is 3/1

6 0

2 years ago