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2 years ago

Please answer the picture

Mathematics

2 answers:

diamong [38]2 years ago

5 0

The area of the shape is 230 cm.

dimaraw [331]2 years ago

3 0

The area is equal to 230 CM.
area for rectangle :
area = width times length
area = 7 x 20 = 140CM
area for triangle:
area = 1/2 times base times height
area = 1/2 x 20 x 9 = 90CM

ADD:
140+90 = 230 CM

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LOTS OF BRAINLY POINTS HELP ASAP PLS TYSM

Sedbober [7]

Answer:

3/10

Step-by-step explanation:

So, you already have the experimental probability. The theoretical probability is 5/10, but that's not what you're lookig for.

5 0

2 years ago

Can someone please help me with this asap ... Please and thank you

OLga [1]

Answer:

Step-by-step explanation:

From the question given in the picture,

a). Since, NR bisects a straight angle ∠MNP,

∠MNR ≅ ∠PNR

m∠MNR + m∠PNR = 180°

2(m∠MNR) = 180°

m∠MNR = 90°

Therefore, ∠MNR and ∠PNR are the right angles.

Since, QN divides ∠MNR in two parts,

Therefore, ∠QNR will be an acute angle (less than 90°).

∠MNR + ∠SNR = ∠MNS

90° + ∠SNR = ∠MNS

Therefore, m∠MNS will be more than 90°.

∠MNS will be an obtuse angle (greater than 90°).

(b). Since, NR divides ∠MNP and ∠QNS,

∠MNR ≅ ∠PNR

∠QNR ≅ ∠SNR

∠MNQ ≅ ∠PNS

(c). m∠MNR = 90°

Since, NR bisects ∠QNS,

∠QNR ≅ ∠RNS

m∠QNR = m∠RNS = 30°

m∠QNR + m∠RNS = 30° + 30°

m∠QNR + m∠RNS = 60°

m∠QNS = 60° [Since, m∠QNS = m∠QNR + m∠RNS]

m∠QNP = m∠QNS + m∠SNP

m∠QNP = m∠QNS + (m∠PNR - m∠SNR)

m∠QNP = 60° + (90° - 30°) = 120°

4 0

2 years ago

Solve the following equations by factorisation method.Only factorisation not dharacharya.

Luba_88 [7]

Hello, please consider the following.

When $x_1$ and $x_2$ are two roots, we can factorise as

$ax^2+bx+c=a(x-x_1)(x-x_2)=a(x^2-(x_1+x_2)x+x_1x_2)=0$

So for the first equation, we can say that the sum of the zeros is

$\dfrac{a^2+b^2}{2}=\dfrac{a^2}{2}+\dfrac{b^2}{2}$

and the product is

$\dfrac{a^2b^2}{4}=\dfrac{a^2}{2}\dfrac{b^2}{2}$

So we can factorise as below.

$4x^2-2(a^2+b^2)x+a^2b^2=(2x-a^2)(2x-b^2)=0$

And the solutions are

$\boxed{\sf \n\bf \ \dfrac{a^2}{2} \ \ and \ \ \dfrac{b^2}{2}}$

For the second equation, we will complete the square and put the constant on the right side and take the root.

Let's do it!

$9x^2-9(a+b)x+2a^2+5ab+2b^2=0\\\\9(x-\dfrac{a+b}{2})^2-9\dfrac{(a+b)^2}{4}+2a^2+5ab+2b^2=0\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{9(a+b)^2-8a^2-20ab-8b^2}{4}\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{9a^2+18ab+9b^2-8a^2-20ab-8b^2}{4}\\\\9(x-\dfrac{a+b}{2})^2=\dfrac{a^2+b^2-2ab}{4}=\dfrac{(a-b)^2}{4}\\\\(x-\dfrac{a+b}{2})^2=\dfrac{(a-b)^2}{2^2\cdot 3^2}$

We take the root, and we find the two solutions

$\begin{aligned}x_1&=\dfrac{a+b}{2}-\dfrac{a-b}{6}\\\\&=\dfrac{3a+3b-a+b}{6}\\\\&=\dfrac{2a+4b}{6}\\\\&\boxed{=\dfrac{a+2b}{3}}\end{aligned}$

$\begin{aligned}x_1&=\dfrac{a+b}{2}+\dfrac{a-b}{6}\\\\&=\dfrac{3a+3b+a-b}{6}\\\\&=\dfrac{4a+2b}{6}\\\\&\boxed{=\dfrac{2a+b}{3}}\end{aligned}$

Thank you.

6 0

3 years ago

At a state fair, every go-kart goes at most 40 mi/h.

Volgvan

Answer: s _< 40

Step-by-step explanation:

7 0

3 years ago

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The legs of a right triangle are 6 units and 7 units. What is the length of the hypotenuse?

UNO [17]

This requires the pythagorean theorem :
a^2 + b^2 = c^2...where a and b are the legs and c is the hypotenuse..and by the way, this can only be used on right triangles

a^2 + b^2 = c^2
6^2 + 7^2 = c^2
36 + 49 = c^2
85 = c^2...take the sqrt of both sides, eliminating the ^2
sqrt 85 = c
9.22 = c....so the hypotenuse is approximately 9.22 units (thats rounded)

5 0

3 years ago

Read 2 more answers