I NEED SOMEONE FROM RSM 7TH GRADE TO COME AND HELP ME ASAP PLZ!!!!!!!!!!!!

Use the graph of the polynomial function to find the factored form of the

irina [24]

Answer:

The answer is

(x-1) (x-9)

8 0

3 years ago

Determine the value of y is x is -10: y=x-91+2

cestrela7 [59]

Answer:

$y = x - 91 + 2 \\ x = - 10 \\ y = - 10 - 91 + 2 \\ y = - 101 + 2 \\ x = - 99$

3 0

3 years ago

Bill got a bank statement and found that his statement balance did not agree with his checkbook

Ivenika [448]

Answer:

write the balance from the bank statement in his checkbook

Step-by-step explanation:

4 0

3 years ago

Need to pick three options. please need help on this

disa [49]

The 4th one is the answer

3 0

3 years ago

Hello help me with this question thanks in advance

Ede4ka [16]

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

<h3>Answer 11 :-</h3>

We have, 

$\sf{HM = 5 cm }$

In square all sides of squares are equal

The perimeter of square 

$\sf{ = 4 × side }$

$\sf{ = 4 × 5 }$

$\sf{ = 20 cm }$

Thus, The perimeter of square is 20 cm

Hence, Option C is correct .

<h3>Answer 12 :-</h3>

We have, 

$\sf{MX = 3.5 cm }$

In square, diagonals are equal and bisect each other at 90°

Here, 

$\sf{MX = MT/2}$

$\sf{MT = 2 * 3.5 }$

$\sf{MT = 7 cm}$

Thus, The MT is 7cm long

Hence, Option C is correct .

<h3>Answer 13 :- </h3>

We have to find the measure of Angle MAT

All angles of square are 90° each

From above

$\sf{\angle{MAT = 90° }}$

Thus, Angle MAT is 90°

Hence, Option B is correct .

<h3>Answer 14 :-</h3>

We know that, 

All the angles of square are equal and 90° each

Therefore, 

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

$\sf{\angle{MHA = 90°/2}}$

$\sf{\angle {MHA = 45°}}$

Thus, Angle MHA is 45°

Hence, Option A is correct

<h3>Answer 15 :- </h3>

Refer the above attachment for solution

Hence, Option A is correct

<h3>Answer 16 :- </h3>

Both a and b

The median of isosceles trapezoid is parallel to the base
The diagonals are congruent

Hence, Option C is correct

<h3>Answer 17 :-</h3>

In rhombus PALM,

All sides and opposite angles are equal

Let O be the midpoint of Rhombus PALM

In ΔOLM, By using Angle sum property :-

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

$\sf{\angle{OLM = 180° - 125°}}$

$\sf{\angle{ OLM = 55° }}$

Now, 

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

OL is the bisector of diagonal AM

Therefore, 

$\sf{\angle{ PLA = 55° }}$

Thus, Angle PLA is 55° .

Hence, Option C is correct

8 0

3 years ago