Marianne owes $56 for tickets to the school play. If each ticket cost $4, how many tickets did she buy?

A discount of 18% on a tennis racquet reduced its price by $16.91. What was the sale price?

Anni [7]

18%=$16.91
1%=$16.91÷18
=$0.93944(3.s.f)
100%=$0.93944x100
=$93.94(2.d.p)

8 0

2 years ago

the following table gives the number of people on the titanic by class and whether or not they survived gi

ziro4ka [17]

Answer:

Where’s the rest of the question ??

Step-by-step explanation:

4 0

2 years ago

Significant figures in a measurement include all of the digits that are

bonufazy [111]

If you were to make your own measurements, your significant digits should include all of the measurable digits (the digits that correspond to the marks on the ruler) as well as one estimated position beyond the smallest measureable digit (the 5 in 3.5 cm, and the 2 in 3.52 cm).

5 0

3 years ago

A blue whale calf weighed 2725 kilograms at birth. A blue whale calf gains 90 kilograms of weight each day for the first 240 day

Ivanshal [37]

Answer:the weight after 225 days is

22885 kilograms

Step-by-step explanation:

The initial weight of the blue whale calf at birth is 2725 kilograms. blue whale calf gains 90 kilograms of weight each day for the first 240 days after its birth. The weight increases in arithmetic progression. This means that the first term of the sequence, a is 2725, the common difference, d is 90.

The formula for the nth term of an arithmetic sequence is expressed as

Tn = a + (n - 1)d

Where

n is the number of terms of the sequence.

a is the first term

d is the common difference

We want to determine its weight, T225 after 225 days after it’s birth. It means that n = 225

Therefore

T225 = 2725 + (225 - 1)90

T225 = 2725 + 224×90 = 2725 + 20160

T225 = 22885

8 0

3 years ago

What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Arte-miy333 [17]

Answer:

The quotient of this division is $(4x^2 -12x + 38)$ . The remainder here would be $-26$ .

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$ .

The divisor $x + 3$ is a polynomial, also about $x$ , but with degree $1$ .

By the division algorithm, the quotient should be of degree $3 - 1 = 2$ , while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$ , $b$ , and $c$ .

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$ .

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$ . On the right-hand side, that would be $a\, x^3$ . Hence $a = 4$ .

Now, given that $a = 4$ , rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$ .

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$ .

The term with a degree of two on the left-hand side has coefficient $(-12)$ . Since the only term on the right hand side with degree two would have coefficient $b$ , $b = -12$ .

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$ .

Subtract $-12x^2 -36x$ from both sides of the equation:

$38x + 7 = c(x + 3)$ .

By the same logic, $c = 38$ .

Hence the quotient would be $(4x^2 - 12x + 38)$ .

6 0

3 years ago