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3 years ago

Write <, >, or = draw a quick picture if you need to.

Mathematics

2 answers:

ser-zykov [4K]3 years ago

6 0

Answer:

Answer is attached to the picture

hope this helps

brainliest appreciated

good luck! have a nice day!

Mashutka [201]3 years ago

3 0

Look at the attached picture

Hope it will help you.

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Being timed help

Makovka662 [10]

I'm assuming 43 is actually 4/3
see what number the previous term must be multiplied by to get the next term
4 *something = 12, 12*something=36
common ratio is 3

6 0

4 years ago

Alguem pra me ajudar na minha prova de matematica daqui 10 min? 8 ano

o-na [289]

Noenoenoenoenoenoenoenoenoe

3 0

3 years ago

find the area of the trapezium whose parallel sides are 25 cm and 13 cm The Other sides of a Trapezium are 15 cm and 15 CM

Snezhnost [94]

$\huge\underline{\red{A}\blue{n}\pink{s}\purple{w}\orange{e}\green{r} -}$

Given - A trapezium ABCD with non parallel sides of measure 15 cm each ! along , the parallel sides are of measure 13 cm and 25 cm

To find - Area of trapezium

Refer the figure attached ~

In the given figure ,

AB = 25 cm

BC = AD = 15 cm

CD = 13 cm

Construction -

$draw \: CE \: \parallel \: AD \: \\ and \: CD \: \perp \: AE$

Now , we can clearly see that AECD is a parallelogram !

$\therefore$ AE = CD = 13 cm

Now ,

$AB = AE + BE \\\implies \: BE =AB - AE \\ \implies \: BE = 25 - 13 \\ \implies \: BE = 12 \: cm$

Now , In ∆ BCE ,

$semi \: perimeter \: (s) = \frac{15 + 15 + 12}{2} \\ \\ \implies \: s = \frac{42}{2} = 21 \: cm$

Now , by Heron's formula

$area \: of \: \triangle \: BCE = \sqrt{s(s - a)(s - b)(s - c)} \\ \implies \sqrt{21(21 - 15)(21 - 15)(21 - 12)} \\ \implies \: 21 \times 6 \times 6 \times 9 \\ \implies \: 12 \sqrt{21} \: cm {}^{2}$

Also ,

$area \: of \: \triangle \: = \frac{1}{2} \times base \times height \\ \\\implies 18 \sqrt{21} = \: \frac{1}{\cancel2} \times \cancel12 \times height \\ \\ \implies \: 18 \sqrt{21} = 6 \times height \\ \\ \implies \: height = \frac{\cancel{18} \sqrt{21} }{ \cancel 6} \\ \\ \implies \: height = 3 \sqrt{21} \: cm {}^{2}$

Since we've obtained the height now , we can easily find out the area of trapezium !

$Area \: of \: trapezium = \frac{1}{2} \times(sum \: of \:parallel \: sides) \times height \\ \\ \implies \: \frac{1}{2} \times (25 + 13) \times 3 \sqrt{21} \\ \\ \implies \: \frac{1}{\cancel2} \times \cancel{38 }\times 3 \sqrt{21} \\ \\ \implies \: 19 \times 3 \sqrt{21} \: cm {}^{2} \\ \\ \implies \: 57 \sqrt{21} \: cm {}^{2}$

hope helpful :D

6 0

2 years ago

Pleaseee help quickly !!!!

lisabon 2012 [21]

Let's solve your equation step-by-step.

(− 5 /8 )(x)=−160

Step 1: Simplify both sides of the equation.

−5 /8

x=−160

Step 2: Multiply both sides by 8/(-5).

( 8 /−5 )x( −5 /8 x)=( 8/ −5 )*(−160)

x=256

Answer:

x=256

8 0

3 years ago

PLEASE HELP !! ILL GIVE BRAINLIEST !!

bearhunter [10]

Answer:

(a) ∠HIK and ∠GFD

Step-by-step explanation:

In this geometry, there are only two pairs of alternate exterior angles:

∠HIK and ∠GFD

∠JIK and ∠EFD

Of these, only the first pair is among the answer choices.

Alternate exterior angles are ones that are outside the parallel lines, and on opposite sides of the transversal.

7 0

3 years ago