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3 years ago

What is the distance between the points (-3,4) and (5,4)?

Mathematics

2 answers:

butalik [34]3 years ago

6 0

Answer:

8 units

Step-by-step explanation:

i hope this helped

marta [7]3 years ago

5 0

Answer:

8/1

Step-by-step explanation:

You have to go up 8 more to get to (5,4) I hope Im right and this helps

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A circle has a radius of 5 in. A central angle that measures 150° cuts off an arc.

Slav-nsk [51]

Answer:

Part 1) The exact value of the arc length is $\frac{25}{6}\pi \ in$

Part 2) The approximate value of the arc length is $13.1\ in$

Step-by-step explanation:

step 1

Find the circumference of the circle

The circumference of a circle is equal to

$C=2\pi r$

we have

$r=5\ in$

substitute

$C=2\pi (5)$

$C=10\pi\ in$

step 2

Find the exact value of the arc length by a central angle of 150 degrees

Remember that the circumference of a circle subtends a central angle of 360 degrees

by proportion

$\frac{10\pi}{360} =\frac{x}{150}\\ \\x=10\pi *150/360\\ \\x=\frac{25}{6}\pi \ in$

step 3

Find the approximate value of the arc length

To find the approximate value, assume

$\pi =3.14$

substitute

$\frac{25}{6}(3.14)=13.1\ in$

7 0

3 years ago

Find the value of x in the isosceles triangle shown below.

Reptile [31]

Answer:

x=10

Step-by-step explanation:

We can use the Pythagorean theorem to find x

The height meets the base at a right angle

The base is bisected ( cut in half) so the base is 6 the height is 8

a^2 + b^2 = c^2

6^2 + 8^2 = x^2

36+64 = x^2

100 = x^2

Taking the square root of each side

sqrt(100) = sqrt(x^2)

10 = x

4 0

3 years ago

Read 2 more answers

21. In 4 + In(4x - 15) = In(5x + 19)

Alexxx [7]

Answer:

$x=-30;\quad \:I\ne \:0$

Step-by-step explanation:

$In\cdot \:4+In\left(4x-15\right)=In\left(5x+19\right)$

$\:4+In\left(4x-15\right):\quad -11nI+4nxI\\In\cdot \:4+In\left(4x-15\right)\\=4nI+nI\left(4x-15\right)\\\\\:In\left(4x-15\right):\quad 4nxI-15nI\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b-c\right)=ab-ac\\a=In,\:b=4x,\:c=15\\=In\cdot \:4x-In\cdot \:15\\=4nxI-15nI\\=In\cdot \:4+4nxI-15nI\\\mathrm{Simplify}\:In\cdot \:4+4nxI-15nI:\quad -11nI+4nxI\\In\cdot \:4+4nxI-15nI\\\mathrm{Group\:like\:terms}\\=4nI-15nI+4nxI\\\mathrm{Add\:similar\:elements:}\:4nI-15nI=-11nI\\=-11nI+4nxI\\$

$\mathrm{Expand\:}In\left(5x+19\right):\quad 5nxI+19nI\\In\left(5x+19\right)\\=nI\left(5x+19\right)\\\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac\\a=In,\:b=5x,\:c=19\\=In\cdot \:5x+In\cdot \:19\\=5nxI+19nI\\\\-11nI+4nxI=5nxI+19nI\\\\\mathrm{Add\:}11nI\mathrm{\:to\:both\:sides}\\-11nI+4nxI+11nI=5nxI+19nI+11nI\\Simplify\\4nxI=5nxI+30nI\\\mathrm{Subtract\:}5nxI\mathrm{\:from\:both\:sides}\\4nxI-5nxI=5nxI+30nI-5nxI\\\mathrm{Simplify}\\-nxI=30nI\\$

$\mathrm{Divide\:both\:sides\:by\:}-nI;\quad \:I\ne \:0\\\frac{-nxI}{-nI}=\frac{30nI}{-nI};\quad \:I\ne \:0\\\mathrm{Simplify}\\\frac{-nxI}{-nI}=\frac{30nI}{-nI}\\\mathrm{Simplify\:}\frac{-nxI}{-nI}:\quad x\\\frac{-nxI}{-nI}\\\mathrm{Apply\:the\:fraction\:rule}:\quad \frac{-a}{-b}=\frac{a}{b}\\=\frac{nxI}{nI}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{xI}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=x\\\mathrm{Simplify\:}\frac{30nI}{-nI}:\quad -30\\\mathrm{Apply\:the\:fraction\:rule}:$

$\quad \frac{a}{-b}=-\frac{a}{b}\\\mathrm{Cancel\:the\:common\:factor:}\:n\\=\frac{30I}{I}\\\mathrm{Cancel\:the\:common\:factor:}\:I\\=-30\\x=-30;\quad \:I\ne \:0$