Ask question

Ask question

All categories

2 years ago

14

CORINA INDRAGESTE FOARTE MULT ANIMALELE IN DAR DE ZIU EI A PRIMIT O CARTE DESPRE LUMEA LOR MINUNATA IN PRIMA ZI EA A CITIT 36 DE

PAGINI IAR IN A DOUA ZI A TREIA ZI DIN REST CATE PAGIN MAI ARE DE CITIT STIIND CA ALBUMUL ARE IN TOTAL 159DE PAGINI?

1 answer:

aliya0001 [1]2 years ago

7 0

Paging that’s what I got for the answer

You might be interested in

The tax rate as a percent, r, charged on an item can be determined using the formula StartFraction c Over p EndFraction minus 1

Nataliya [291]

Answer:

43.20

Step-by-step explanation:

Had to take one for the team

4 0

1 year ago

Read 2 more answers

At what points does the helix r(t) = sin t, cos t, t intersect the sphere x2 + y2 + z2 = 65? (round your answers to three decima

Firdavs [7]

$\mathbf r(t)=\langle x(t),y(t),z(t)\rangle=\langle\sin t,\cos t,t\rangle$

$x^2+y^2+z^2=\sin^2t+\cos^2t+t^2=65$
$\implies t^2=64$
$\implies t=\pm8$

7 0

1 year ago

∠E and ∠F are vertical angles with m∠E=5x+10 and m∠F=7x−12 .<br><br> What is the value of x?

In-s [12.5K]

Vertical angles are equal...so set ur angles equal to each other and solve for x

5x + 10 = 7x - 12
12 + 10 = 7x - 5x
22 = 2x
22/2 = x
11 = x <==

6 0

2 years ago

If f(1) = 0 what are all the roots of the function f(x)=x^3+3x^2-x-3 use the remainder theorem.

ArbitrLikvidat [17]

Solution:

As we are given that f(1) = 0 .

It mean that $(x-1)$ is one of the factor of the given equation.

Remainder theorem can be applied as below:

$\frac{(x^3+3x^2-x-3)}{(x-1)}=\frac{x^3-x^2+4x^2-4x+3x-3}{(x-1)}\\ \\\frac{x^3-x^2+4x^2-4x+3x-3}{(x-1)}=\frac{x^2(x-1)+4x(x-1)+3(x-1)}{(x-1)} \\\\\frac{x^2(x-1)+4x(x-1)+3(x-1)}{(x-1)}=\frac{(x^2+4x+3)(x-1)}{(x-1)} \\\\\frac{(x^2+4x+3)(x-1)}{(x-1)} =\frac{(x^2+3x+x+3)(x-1)}{(x-1)} \\\\\frac{(x^2+3x+x+3)(x-1)}{(x-1)} =\frac{(x-1)(x+3)(x+1)}{(x-1)}$

Hence the factors are (x-1),(x+3) and (x+1).

Hence the correct option is B.

5 0

2 years ago

Read 2 more answers

A country's population in 1993 was 94 million. in 1999 in was 99 million. estimate the population in 2005 using the exponential

Jet001 [13]

The initial population is
P₀ = 94 million in 1993

The growth formula is
$P(t) = P_{0}e^{kt}$
where P(t) is the population (in millions) after t years, measured from 1993.
k = constant.

Because P(5) = 99 million (in 1999),
$94e^{5k} = 99 \\e^{5k}=1.0532 \\ 5k = ln(1.0532) \\ k = 0.010367$

In the year 2005, t = 12 years, and
$P(12)=94e^{0.010367*12} = 106.45$

Answer: 106 million (nearest million)

7 0

1 year ago