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3 years ago

I need help with Math ASAP plz !!

Mathematics

1 answer:

Elina [12.6K]3 years ago

6 0

Answer: 1 square and 4 triangles

Step-by-step explanation:

you can see the square at the bottom. Around you can see 1 triangle in the front 2 on the sides and one in the back.

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In the diagram abc=adb=90, ad=p and dc=q. Use similar triangles to show that x2=pz<br> plzz anyoneee

kramer

Answer:

By comparing the ratios of sides in similar triangles ΔABC and ΔADB,we can say that $x^{2} =pz$

Step-by-step explanation:

Given that ∠ABC=∠ADC, AD=p and DC=q.

Let us take compare Δ ABC and Δ ADB in the attached file , ∠A is common in both triangles

and given ∠ABC=∠ADB=90°

Hence using AA postulate, ΔABC ≈ ΔADB.

Now we will equate respective side ratios in both triangles.

$\frac{AB}{AC}= \frac{AD}{AB}=\frac{BD}{BC}$

Since we don't know BD , BC let us take first equality and plugin the variables given in respective sides.

$\frac{x}{z}= \frac{p}{x}$

Cross multiply

$x^{2}=pz$

Hence proved.

7 0

3 years ago

NO LINKS OR ANSWERING QUESTIONS YOU DON'T KNOW!!! THIS IS NOT A TEST OR AN ASSESSMENT!! Please help me with these math questions

Art [367]

Answer:

See below

Step-by-step explanation:

3. What are two ways that a vector can be represented?

Considering a vector $\vec{v}$ in some vector space $\mathbb R^n$ we have

$\vec{v} = \langle a,b\rangle$

This is the component form. I don't like that way. It is probably used in high school, but

$\vec{v} = \begin{pmatrix} a\\ b\\ \end{pmatrix}$

is preferable because the inner product on $\mathbb R^n$ is defined to be

$\langle a,b\rangle := \sum_{i = 1}^n a_i b_i$

You can also write it using linear form such as $\vec{v} = 2i+2j$

For this question, I think you meant

vectors

$\vec{u_1} = (-8, 12)$

$\vec{u_2} = (13, 15)$

Once

$\cos(\theta)=\dfrac{\vec{u_1} \cdot\vec{u_2}}{||\vec{u_1}||||\vec{u_2}||}$

Considering that the dot product is

$\vec{u_1}\cdot \vec{u_2} = (-8)\cdot 13 + 12\cdot 15 = -104+180= 76$

and the norm of $\vec{u_1}$ is $||\vec{u_1}|| = \sqrt{(-8)^2 + 12^2} = \sqrt{64 + 144}= \sqrt{208}$

and the norm of $\vec{u_2}$ is $||\vec{u_2}|| = \sqrt{13^2 + 15^2} = \sqrt{169 + 225}= \sqrt{394}$

Thus,

$\cos(\theta)=\dfrac{76}{\sqrt{208} \sqrt{394}} = \dfrac{19}{\sqrt{13}\sqrt{394}}=\dfrac{19}{\sqrt{5122}}$

$\therefore \theta = \arccos \left(\dfrac{19}{\sqrt{5122}} \right)$

3 0

2 years ago

I need help please):

Neko [114]

Answer:

7) x = 5

8) x = 10

Step-by-step explanation:

7) 2x-3=x+2

x=5

8) 4x=2x+20

2x=20

x=10

7 0

3 years ago

What is the slope and y-intercept this table?

svlad2 [7]

Slope=10

y intercept=17

4 0

3 years ago

Read 2 more answers

____ [38]

Answer:

y = 5x+3

Step-by-step explanation:

a general line is represented by the equation:

y = mx + n

a line with m=5, that goes through (-1,-2), should satisfy the equation:

-2 = 5×(-1) + n

this is because any point on the line, satisfies it's equation.

from this we get n = 3, hence our answer

4 0

2 years ago