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4 years ago

Round 43.586 to nearest tenth

Mathematics

2 answers:

ruslelena [56]4 years ago

5 0

The answer is 43.6 for the question

MArishka [77]4 years ago

5 0

43.6 because 8 is greater than 4 so you round up

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A camera regularly priced at 295 was placed on sale at 236. What percent of the regular price was the sale price?

skad [1K]

Answer:

80%

Step-by-step explanation:

regular price was given as $295.

The sale of the camera was $236.

Needed Percentage of the regular price =( 236/295)

= 0.8

=(0.8 × 100%)

= 80%

6 0

3 years ago

What would this be?:)

Mnenie [13.5K]

Answer:

Step-by-step explanation:

3 0

3 years ago

Hi please help , With solution if it's ok

Snowcat [4.5K]

Answers with Method:-

1) Multiply the length value by 100,000.

$2.5 \: hm = 2.5 \times 100000$

$= >2.5 \: hm = 250000 \: mm$

2) For this, divide the length value by 1000.

$= > \frac{1800}{1000}$

$= > 1.8 \: dam$

3) For finding the approximate value, just multiply the value of length by 1.609.

$= >6450 \: m = (6450 \times 1.609) \: km$

$= > 10380.27 \: km$

4) Multiply the given mass value by 100.

$= > 1.2 \: kg = 1.2 \times 100$

$= > 1200 \: g$

5) Multiply the given value by 10,000.

$4.37 \: dag = 4.37 \times 10000$

$= > 43700 \: mg$

5 0

2 years ago

Use the the triangle.

KATRIN_1 [288]

For a 45 45 90 triangle if the legs are x length then the hyptonuse is x√2

a.
7=x
so hyptonuse=7√2 in

b.
sin(45°)=(√2)/2

c.
not sure what you need SAS for
but, area=1/2 times base times height
base=height=7
area=1/2 times 7 times 7=24.5 square inches

d. not sure, same as above?

7 0

3 years ago

Use the quadratic formula to solve the following equation -3x^2-x-3=0

tigry1 [53]

Answer:

$x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i$ are two roots of equation $-3 x^{2}-x-3=0$

Solution:

Need to solve given equation using quadratic formula.

$-3 x^{2}-x-3=0$

General form of quadratic equation is $a x^{2}+b x+c=0$

And quadratic formula for getting roots of quadratic equation is

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

In our case b = -1 , a = -3 and c = -3

Calculating roots of the equation we get

$\begin{array}{l}{x=\frac{-(-1) \pm \sqrt{(-1)^{2}-4(-3)(-3)}}{2 \times-3}} \\\\ {x=-\frac{1}{6} \pm\left(-\frac{\sqrt{-35}}{6}\right)}\end{array}$

Since $b^{2}-4 a c$ is equal to -35, which is less than zero, so given equation will not have real roots and have complex roots.

$\begin{array}{l}{\text { Hence } x=-\frac{1}{6}-\frac{\sqrt{35}}{6} i \text { and } x=-\frac{1}{6}+\frac{\sqrt{35}}{6} i \text { are two roots of equation - }} \\ {3 x^{2}-x-3=0}\end{array}$

8 0

4 years ago