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mel-nik [20]
3 years ago
13

What is the slope of the line x=5

Mathematics
1 answer:
Andre45 [30]3 years ago
5 0
The slope would go vertically starting it at 5. you place your dot at 5 and draw a line going up and down throughout both the top and bottom of the right side of the graph. hope it helped
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Help please!<br> Much appreciated
Mamont248 [21]
Hello, this is a volume question because it involves 3 measures, the length width and height, so what we would do is we would multiply, 10 x 6 x 8 which is 48
7 0
2 years ago
Read 2 more answers
Find the distance between a point (– 2, 3 – 4) and its image on the plane x+y+z=3 measured parallel to a line
Ber [7]

Answer:

The distance is:  

\displaystyle\frac{3\sqrt{142}}{10}

Step-by-step explanation:

We re-write the equation of the line in the format:

\displaystyle\frac{x+2}{3}=\frac{y+\frac{3}{2}}{2}=\frac{z+\frac{4}{3}}{\frac{5}{3}}

Notice we divided the fraction of y by 2/2, and the fraction of z by 3/3.

In that equation, the director vector of the line is built with the denominators of the equation of the line, thus:

\displaystyle\vec{v}=\left< 3, 2, \frac{5}{3}\right>

Then the parametric equations of the line along that vector and passing through the point (-2, 3, -4) are:

x=-2+3t\\y=3+2t\\\displaystyle z=-4+\frac{5}{3}t

We plug them into the equation of the plane to get the intersection of that line and the plane, since that intersection is the image on the plane of the point (-2, 3, -4)  parallel to the given line:

\displaystyle x+y+z=3\to -2+3t+3+2t-4+\frac{5}{3}t=3

Then we solve that equation for t, to get:

\displaystyle \frac{20}{3}t-3=3\to t=\frac{9}{10}

Then plugging that value of t into the parametric equations of the line we get the coordinates of the intersection:

\displaystyle x=-2+3\left(\frac{9}{10}\right)=\frac{7}{10}\\\displaystyle y=3+2\left(\frac{9}{10}\right)=\frac{24}{5} \\\displaystyle z=-4+\frac{5}{3}\left(\frac{9}{10}\right)=-\frac{5}{2}

Then to find the distance we just use the distance formula:

\displaystyle d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2+(z_1-z_2)^2}

So we get:

\displaystyle d=\sqrt{\left(-2-\frac{7}{10}\right)^2+\left(3-\frac{24}{5}\right)^2+\left(-4 +\frac{5}{2}\right)^2}=\frac{3\sqrt{142}}{10}

3 0
3 years ago
Hello help me with this question thanks in advance​
Ede4ka [16]

\bold{\huge{\green{\underline{ Solutions }}}}

<h3><u>Answer </u><u>1</u><u>1</u><u> </u><u>:</u><u>-</u></h3>

<u>We </u><u>have</u><u>, </u>

\sf{HM = 5 cm }

  • <u>In </u><u>square </u><u>all </u><u>sides </u><u>of </u><u>squares </u><u>are </u><u>equal </u>

<u>The </u><u>perimeter </u><u>of </u><u>square </u>

\sf{ = 4 × side }

\sf{ = 4 × 5 }

\sf{ = 20 cm }

Thus, The perimeter of square is 20 cm

Hence, Option C is correct .

<h3><u>Answer </u><u>1</u><u>2</u><u> </u><u>:</u><u>-</u></h3>

<u>We </u><u>have</u><u>, </u>

\sf{MX  = 3.5 cm }

  • <u>In </u><u>square</u><u>,</u><u> </u><u>diagonals </u><u>are </u><u>equal </u><u>and </u><u>bisect </u><u>each </u><u>other </u><u>at </u><u>9</u><u>0</u><u>°</u>

<u>Here</u><u>, </u>

\sf{MX  = MT/2}

\sf{MT = 2 * 3.5 }

\sf{MT = 7 cm}

Thus, The MT is 7cm long

Hence, Option C is correct .

<h3><u>Answer </u><u>1</u><u>3</u><u> </u><u>:</u><u>-</u><u> </u></h3>

<u>We </u><u>have </u><u>to </u><u>find </u><u>the </u><u>measure </u><u>of </u><u>Ang</u><u>l</u><u>e</u><u> </u><u>MAT</u>

  • <u>All </u><u>angles </u><u>of </u><u>square </u><u>are </u><u>9</u><u>0</u><u>°</u><u> </u><u>each </u>

<u>From </u><u>above </u>

\sf{\angle{MAT  = 90° }}

Thus, Angle MAT is 90°

Hence, Option B is correct .

<h3><u>Answer </u><u>1</u><u>4</u><u> </u><u>:</u><u>-</u></h3>

<u>We </u><u>know </u><u>that</u><u>, </u>

  • <u>All </u><u>the </u><u>angles </u><u>of </u><u>square </u><u>are </u><u>equal </u><u>and </u><u>9</u><u>0</u><u>°</u><u> </u><u>each </u>

<u>Therefore</u><u>, </u>

\sf{\angle{MHA  = }}{\sf{\angle{ MHT/2}}}

\sf{\angle{MHA = 90°/2}}

\sf{\angle {MHA = 45°}}

Thus, Angle MHA is 45°

Hence, Option A is correct

<h3><u>Answer </u><u>1</u><u>5</u><u> </u><u>:</u><u>-</u><u> </u></h3>

Refer the above attachment for solution

Hence, Option A is correct

<h3><u>Answer </u><u>1</u><u>6</u><u> </u><u>:</u><u>-</u><u> </u></h3>

Both a and b

  • <u>The </u><u>median </u><u>of </u><u>isosceles </u><u>trapezoid </u><u>is </u><u>parallel </u><u>to </u><u>the </u><u>base</u>
  • <u>The </u><u>diagonals </u><u>are </u><u>congruent </u>

Hence, Option C is correct

<h3><u>Answer </u><u>1</u><u>7</u><u> </u><u>:</u><u>-</u></h3>

In rhombus PALM,

  • <u>All </u><u>sides </u><u>and </u><u>opposite </u><u>angles </u><u>are </u><u>equal </u>

Let O be the midpoint of Rhombus PALM

<u>In </u><u>Δ</u><u>OLM</u><u>, </u><u>By </u><u>using </u><u>Angle </u><u>sum </u><u>property </u><u>:</u><u>-</u>

\sf{35° + 90° + }{\sf{\angle{ OLM = 180°}}}

\sf{\angle{OLM = 180° - 125°}}

\sf{\angle{ OLM = 55° }}

<u>Now</u><u>, </u>

\sf{\angle{OLM = }}{\sf{\angle{OLA}}}

  • <u>OL </u><u>is </u><u>the </u><u>bisector </u><u>of </u><u>diagonal </u><u>AM</u>

<u>Therefore</u><u>, </u>

\sf{\angle{ PLA = 55° }}

Thus, Angle PLA is 55° .

Hence, Option C is correct

8 0
2 years ago
The sum of two prime numbers is 85. What is the product of these two prime numbers?
hodyreva [135]
Observe that 85 is an odd number. That means, one of the prime numbers must be even to have an odd sum with another odd number. As we know, the only even prime number is 2. Hence, the other number is 83. So, the product of these two prime numbers is 83 x 2 = 166. Thus, the answer is 166<span>. </span>
6 0
3 years ago
A post 8 feet tall casts a shadow 12 feet long at the same time that a flagpole casts a shadow 96 feet long, how tall is the fla
lara [203]

Answer:

64 ft

Step-by-step explanation:

The shadow of the post is 12 feet long. The shadow of the flagpole is 96 feet long.

The shadow of the flagpole is 8 times longer than the shadow of the post.

8 x 8 = 64

8 0
3 years ago
Read 2 more answers
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