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4 years ago

8

The african bush elephant weighs between 4.4 tons and 7.7 tons. What are its least and greatest wieghts rounded to the nearest t

on

2 answers:

VashaNatasha [74]4 years ago

7 0

4.4 rounds to 4.

7.7 rounds to i

WARRIOR [948]4 years ago

6 0

4.4=4 tons rounded

7.7=8 tons rounded

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Charles is reading about computers. He learns that a computer processor can perform one command in approximately 0.000000016 n

Andru [333]

To convert our number to scientific notation, we are going to move the decimal point to the right until we have only one non-zero digit to the left of the decimal point, and then, we are going to multiply that number by a negative power of 10 equal to the number of times we moved the decimal point to the right:
<span>
Our number is 0.000000016, so we need to move the decimal point 8 times to the right to have </span>only one non-zero digit to the left of the decimal point (1.6). Since we moved the decimal point eight times to the right, we are going to multiply 1.6 by y a negative power of 10 equal to the number of times we moved the decimal point to the right i.e. $10^{-8}$ .
So 0.000000016 expressed in scientific notation is 1.6x $10^-8$ , or in calculator notation 1.6E-8

We can conclude that the correct answers are <span>1.6E-8 and </span><span>1.6 × 10-8</span>

3 0

3 years ago

Read 2 more answers

Which equation could represent the relationship shown in the scatter plot?

Allisa [31]

Y = mx + b, where m is the slope, and b is the y-intercept. If we were to consider a line of best fit that would represent all these points with a linear line, it'd cross the y-axis with a y-value of 10, so we can fill in y = mx + b to y = mx + 10. Now, it runs down from left to right, so we know the slope must be negative. The only equation that fits this description is y= -3/4x + 10

4 0

3 years ago

use the pythagorean theorem to find the missing side length. 2 write sin, cosine, tangent of each angel.

Harlamova29_29 [7]

Answer:

1) $x=5\sqrt{3}$

2) $sin(a)=5\sqrt{3}/14$

$sin(b)=11/14$

$cos(a)=11/14$

$cos(b)=5\sqrt{3}/14$

$tan(a)=5\sqrt{3}/11$

$tan(b)=11/5\sqrt{3}$

Step-by-step explanation:

The Pythagorean Theorem is:

$a^{2}=b^{2}+c^{2}$

Where a is the hypotenuse and b and c are the legs.

The missing side lenght is one of the legs, then you must solve for one of them. Therefore, this is:

$x=\sqrt{(14yd)^{2}-(11yd)^{2}}=5\sqrt{3}yd$

MEASURE OF ANGLE:

Keep the identities on mind:

$sin\alpha=opposite/hypotenuse$

$cos\alpha=adjacent/hypotenuse$

$tan\alpha=opposite/adjacent$

Susbstitute values, then:

$sin(a)=5\sqrt{3}/14$

$sin(b)=11/14$

$cos(a)=11/14$

$cos(b)=5\sqrt{3}/14$

$tan(a)=5\sqrt{3}/11$

$tan(b)=11/5\sqrt{3}$

4 0

4 years ago

Read 2 more answers

If we list all the natural numbers below 20 that are multiples of 7 or 11, we get 7, 11 and 14. The sum of these multiples is 32

velikii [3]

Number of multiples of 7 up to 1337: $\left\lfloor\dfrac{1337}7\right\rfloor=191$
Number of multiples of 11 up to 1337: $\left\lfloor\dfrac{1337}{11}\right\rfloor=121$
Number of multiples of 77 up to 1337: $\left\lfloor\dfrac{1337}{77}\right\rfloor=17$

This means there are $191+121-17=295$ distinct multiples of 7 *or* 11 up to 1337.

The sum of these multiples is

$\displaystyle\sum_{k=1}^{191}7k+\sum_{k=1}^{121}11k-\sum_{k=1}^{17}77k$

which can be computed using the well-known formula,

$\displaystyle\sum_{k=1}^nk=\dfrac{n(n+1)}2$

So you have

$\displaystyle\sum_{k=1}^{191}7k+\sum_{k=1}^{121}11k-\sum_{k=1}^{17}77k=7\dfrac{191\times192}2+11\dfrac{121\times122}2-77\dfrac{17\times18}2=197762$

8 0

3 years ago

Please verify the following trigonometric identities.

Nina [5.8K]

Seems like there is a correction in the first question (RHS is tanx.tany)

(i) For convenience: let tanx = a ; tany = b

Thus, tanx + tany = a + b

Moreover, cotx = 1/tanx = 1/a ; coty = 1/b

Thus,

cotx + coty = 1/a + 1/b = (a + b)/ab

Hence,

=> (tanx + tany)/(cotx + coty)

=> (a + b) / { (a + b)/ab }

=> ab(a + b)/(a + b)

=> ab => tanx.tany , proved.

(ii) For convenience: let sinx = a ; cosx = b

As we know, sin²x + cos²x = 1 => a²+b²=1

=> (a³ + b³)/(a + b)

=> (a + b)(a² + b² - ab) / (a + b)

=> (a² + b² - ab)

=> 1 - ab => 1 - sinx.cosx , proved

(iii): let x/2 = A

=> tan(x/2) + cosx.tan(x/2)

=> tanA + cos2A.tanA

=> tanA [1 + cos2A]

=> tanA (2cos²A) {1+cos2A = 2cos²A}

=> (sinA/cosA) (2cos²A)

=> sinA (2cosA)

=> 2sinAcosA

=> sin2A

=> sin2(x/2)

=> sinx proved

Letting x/2 = A is not mandatory. I did it to decease words*(in a line).

<u>Indentities used</u>:

• sin²A + cos²A = 1

• (a³ + b³) = (a + b)(a² + b² - 1)

• 1 + cosA = 2 cos²(A/2)

• tanA = sinA/cosA.

• 2sinAcosA = sin2A

8 0

3 years ago

Read 2 more answers