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3 years ago

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HELP PLEASE!

1 answer:

Grace [21]3 years ago

8 0

Answer and workings are in the attachments below.

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PLSSSSS HELPOLP URGENT!!!!

murzikaleks [220]

Answer:
Q (4,3)
P (12,1)

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3 years ago

Given the equation Square root of 2x plus 1 = 3, solve for x and identify if it is an extraneous solution

maksim [4K]

To answer the question above, raise both sides of the equation to the power of two.
(sqrt (2x + 1))² = 3²
This will simplify into,
2x + 1 = 9
Solving for the value of x gives an answer of x = 4. This value is not an extraneous root as it satisfy the original equation even if it is inside the radical sign. (I hope this is right)

6 0

3 years ago

Read 2 more answers

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

3 years ago

Find the area of AAOB<br> 9√3 un²<br> 3√3 un²<br> 4.5√3 un

butalik [34]

Answer:

Step-by-step explanation:

The area of the equilateral triangle is $\frac{\sqrt{3}}{4}(6)^{2}=\boxed{9\sqrt{3}}$

7 0

2 years ago

What percentage of the birds shown in the pie chart are notgreen?

Gre4nikov [31]

A 80% would be your answer!

3 0

3 years ago

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