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3 years ago

Please help I’ll give brainliest

Mathematics

1 answer:

ivanzaharov [21]3 years ago

8 0

Answer:

a) 421mm² = 0.000421 m²

b) 9.3kg²= 9,300,000 g²

c) 25kL²= 25,000,000 L²

d) 1439m²= 0.001439 km²

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576 = 96 × 6

aliya0001 [1]

Answer:

Step-by-step explanation

A.96=6+576

B.576=96+6

C. 96=6×576

D. 576=96×6

5 0

3 years ago

It says priya's family exchanged 250 dollars for 4,250 pesos. Riya bought a sweater for 510 pesos. how many did the sweater cost

ratelena [41]

$1 = $20.19 pesos, try dividing 510 pesos by 20.19. That should give you the dollar amount.

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3 years ago

20 increased by twice jennys savings

Rudiy27

20+j*2
Ye, that's about right aye

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4 years ago

HELPSelect the correct answer.<br>Which table shows a proportional relationship between a and b?

romanna [79]

Answer:

Step-by-step explanation:

table B: because when x increases y increases at the same rate and stay the same . the graph has proportional relation when it is a straight line passes through origin

for B :25/20=30/24=40/32=5/4

y=5/4 x

8 0

3 years ago

Read 2 more answers

What are the orders of 3,7,9,11,13,17 and 19(mod20)?does 20 have primitive roots?

bezimeni [28]

$3\equiv3\mod{20}$
$3^2\equiv9\mod{20}$
$3^3\equiv27\equiv7\mod{20}$
$3^4\equiv3\cdot3^3\equiv3\cdot7\equiv21\equiv1\mod{20}$

$7\equiv7\mod{20}$
$7^2\equiv49\equiv9\mod{20}$
$7^3\equiv7\cdot7^2\equiv63\equiv3\mod{20}$
$7^4\equiv7\cdot7^3\equiv21\equiv1\mod{20}$

$9\equiv9\mod{20}$
$9^2\equiv3^4\equiv1\mod{20}$

$11\equiv11\mod{20}$
$11^2\equiv121\equiv1\mod{20}$

$13\equiv-7\equiv13\mod{20}$
$13^2\equiv169\equiv9\mod{20}$
$13^3\equiv13\cdot13^2\equiv(-7)9\equiv-63\equiv-3\mod{20}$
$13^4\equiv13\cdot13^3\equiv(-7)(-3)\equiv21\equiv1\mod{20}$

$17\equiv-3\equiv17\mod{20}$
$17^2\equiv(-3)^2\equiv9\mod{20}$
$17^3\equiv(-3)^3\equiv-27\equiv3\mod{20}$
$17^4\equiv(-3)^4\equiv81\equiv1\mod{20}$

$19\equiv-1\equiv19\mod{20}$
$19^2\equiv19(-1)\equiv-19\equiv1\mod{20}$

Generally speaking, a number $x$ coprime to $n$ will be a primitive root of $n$ if we have $x^n\equiv x\mod{n}$ , or $x^{n-1}\equiv1\mod{n}$ . In other words, if $x$ is of order $n-1$ modulo $n$ , then $x$ is a primitive root of $n$ .

Since none of these numbers has order 19, it follows that 20 does not have any primitive roots.

6 0

4 years ago