Find the missing exponent: 5^11/5?=5^4

Mathematics

1 answer:

Sergio [31]2 years ago

6 0

The missing exponent is 5^7 your welcome

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A rectangle and a triangle with the same base and height have the same area.

OverLord2011 [107]

The answer is false because if you attach to triangles together in a way so that the figure would have 4 straight lines, you will get a rectangle. that means half of a rectangle is a triangle, so the area of both cannot be the same even if they have the same base and height.

for example the base and height of both is 3cm and 5cm.

area of the rectangle then would be,
area= base× height
=3cm×5cm=15cmsquared

then the area of the triangle would be,
area= base×height divided by 2
=3cm×5cm÷2= 7.5cm

so your answer is false
hope this helps....

5 0

3 years ago

Can someone Help please !!!

Sunny_sXe [5.5K]

Answer:

parellel lines have same slope so slope of l2 is same as that of l1

slope of l1 = 5-1/-2-(-3)

4/1

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2 years ago

The temperature Rose 4 degrees each hour for 3 hours what was the total change in the temperature?

ozzi

12 degrees change in temp.

4 0

3 years ago

Find the multiplicative inverse of 6 + 2i

Marina CMI [18]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2774989

________________

Find the multiplicative inverse of

$\mathsf{z=6+2i}$

________

The inverse multiplicative of $\mathsf{z=a+bi}$ is

$\mathsf{\dfrac{1}{z}}\\\\\\
=\mathsf{\dfrac{1}{a+bi}\qquad\quad(a\ne 0~~and~~b\ne 0)}\\\\\\
=\mathsf{\dfrac{1}{a+bi}\cdot \dfrac{a-bi}{a-bi}}\\\\\\
=\mathsf{\dfrac{1\cdot (a-bi)}{(a+bi)\cdot (a-bi)}}\\\\\\
=\mathsf{\dfrac{a-bi}{a^2-\,\diagup\hspace{-10}abi+\,\diagup\hspace{-10}abi-(bi)^2}}$

$=\mathsf{\dfrac{a-bi}{a^2-b^2\cdot i^2}}\\\\\\
=\mathsf{\dfrac{a-bi}{a^2-b^2\cdot (-1)}}\\\\\\
=\mathsf{\dfrac{a-bi}{a^2+b^2}}\\\\\\\\
\therefore~~\mathsf{\dfrac{1}{a+bi}=\dfrac{a}{a^2+b^2}-\dfrac{b}{a^2+b^2}\,i\qquad\quad\checkmark}$

________

For this question,

$\mathsf{z=6+2i}$

So,

$\mathsf{\dfrac{1}{z}}\\\\\\
=\mathsf{\dfrac{1}{6+2i}}\\\\\\
=\mathsf{\dfrac{6}{6^2+2^2}-\dfrac{2}{6^2+2^2}\,i}\\\\\\
=\mathsf{\dfrac{6}{36+4}-\dfrac{2}{36+4}\,i}\\\\\\
=\mathsf{\dfrac{6}{40}-\dfrac{2}{40}\,i}$

$\therefore~~\mathsf{\dfrac{1}{z}=\dfrac{3}{20}-\dfrac{1}{20}\,i}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

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Select the two figures that are similar to each other

Masteriza [31]

The blue one and the small purple one are similar because they have the same angles.

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