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3 years ago

The answer to the length of x is in mm. please help!

Mathematics

1 answer:

romanna [79]3 years ago

4 0

Answer:

3.69 mm

Step-by-step explanation:

So it's a right triangle so you can use a^2+b^2=c^2

Plugging in what we know the equation becomes:

2.8^2+2.4^2=c^2 (C = x because x is the hypotenuse)

7.84+5.76=c^2

13.6=c^2

The square root of 3.68782 which rounds to 3.69

So x=3.69mm

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Which relationship describes angles 1 and 2?

Sholpan [36]

Answer:

vertical angles and complementary

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3 years ago

I need help asap pls and thank you ;)

olga55 [171]

Answer:

$\text{Length of AB is }\frac{ah}{a+h}$

Step-by-step explanation:

Given △KMN, ABCD is a square where KN=a, MP⊥KN, MP=h.

we have to find the length of AB.

Let the side of square i.e AB is x units.

As ADCB is a square ⇒ ∠CDN=90°⇒∠CDP=90°

⇒ CP||MP||AB

In ΔMNP and ΔCND

∠NCD=∠NMP (∵ corresponding angles)

∠NDC=∠NPM (∵ corresponding angles)

By AA similarity rule, ΔMNP~ΔCND

Also, ΔKAP~ΔKPM by similarity rule as above.

Hence, corresponding sides are in proportion

$\frac{ND}{NP}=\frac{CD}{MP} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{AB}{PM} \\\\\frac{ND}{NP}=\frac{x}{h} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{x}{h}\\\\\frac{NP}{ND}=\frac{h}{x} \thinspace\thinspace and\thinspace\thinspace \frac{KP}{KA}=\frac{h}{x}\\\\\frac{PD}{ND}=\frac{h}{x}-1 \thinspace\thinspace and\thinspace\thinspace \frac{AP}{KA}=\frac{h}{x}-1\\$

$KA(\frac{h}{x}-1)=AP$

$ND(\frac{h}{x}-1)=PD$

Adding above two, we get

$(KA+ND)(\frac{h}{x}-1)=(AP+PD)$

⇒ $(KN-AD)=\frac{x}{(\frac{h}{x}-1)}$

⇒ $a-x=\frac{x}{(\frac{h}{x}-1)}$

⇒ $a-x=\frac{x^2}{h-x}$

⇒ $x^2=ah-ax-xh+x^2$

⇒ $x(h+a)=ah$

⇒ $x=\frac{ah}{a+h}$

3 0

3 years ago

Evaluate:

Anni [7]

Answer:

a) = 4.5

b) = 3.3

Step-by-step explanation:

Before solving our problems given to us let us under stand the rule of cube roots

It says $\sqrt[3]{x \times x \times x} = x$ -----(A)

Also

$\sqrt[3]{x \times x \times x \times y \times y \times y} = x \times y$ ---(B)

Now let us see each part one by one

a) we have

$\sqrt[3]{64} + \sqrt[3]{0.027} + \sqrt[3]{0.008}$

Now 64 = 4 x 4 x 4

0.027 = 0.3 x 0.3 x 0.3

0.008 = 0.2 x 0.2 x 0.2

substituting these values

$\sqrt[3]{4 \times 4 \times 4} + \sqrt[3]{0.3 \times 0.3 \times 0.3} + \sqrt[3]{0.2 \times 0.2 \times 0.2}$

Applying Rule A in above

$=4+0.3+0.2$

$=4+0.5$

4.5

b) we have $\sqrt[3]{0.3 \times 0.3 \times 0.3 \times 11 \times 11 \times 11}$

Applying the B rule in this

$=0.3 \times 11$

3.3

4 0

3 years ago

I CANT DLEETE THIS HELP

larisa86 [58]

Answer:

u can not delete it because u already posted it the only way for it to be deleted is if u are reported for what u post then the moderators see it and make it poof

Step-by-step explanation:

5 0

2 years ago

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Ill mark brainlist plss help

Gelneren [198K]

Answer:

quadrilateral, parallelogram, rhombus

8 0

3 years ago

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