All categories

3 years ago

Which number is not in scientific notation? 9 x 10^19 7.8 x 10^6 1.45 x 10^-7 0.33 x 10^-15

Mathematics

2 answers:

Svetach [21]3 years ago

4 0

Answer:

please tell me the answer i need it thank you

Step-by-step explanation:

Len [333]3 years ago

3 0

1.45 x 10^-7 isn't in scientific notation because if it's negative, the decimal starts before the first real number, right?

You might be interested in

2x(5x-2) expanding brackets

Inga [223]

Step-by-step explanation:

2x(5x-2)

= 2x*5x - 2x*2 (Distributive property)

= 10x^2 - 4x

8 0

2 years ago

There are 5 roses in a vaise of 14 flowers. The rest are daisies. (What is the ratio of daisies to roses)

BlackZzzverrR [31]

Answer:

problem a

5 to 9

5/9

problem b

14 to 9

14/9

6 0

3 years ago

Please help me?<br><br> please and thank you!

JulsSmile [24]

Answer:

-1/12

Step-by-step explanation:

change divide sign to multiply by reciprocating

7 0

2 years ago

HELP WITH THESE 3 questions please !!! Please dont give me the wrong answers thanks I really need help

natta225 [31]

Answer:

We use similar triangles to get

brainliest pls

5 0

3 years ago

Evaluate the iterated integral 2 0 2 x sin(y2) dy dx. SOLUTION If we try to evaluate the integral as it stands, we are faced wit

nignag [31]

Answer:

Step-by-step explanation:

Given that:

$\int^2_0 \int^2_x \ sin (y^2) \ dy dx \\ \\ \text{Using backward equation; we have:} \\ \\ \int^2_0\int^2_0 sin(y^2) \ dy \ dx = \int \int_o \ sin(y^2) \ dA \\ \\ where; \\ \\ D= \Big\{ (x,y) | }0 \le x \le 2, x \le y \le 2 \Big\}$

$\text{Sketching this region; the alternative description of D is:} \\ D= \Big\{ (x,y) | }0 \le y \le 2, 0 \le x \le y \Big\}$

$\text{Now, above equation gives room for double integral in reverse order;}$

$\int^2_0 \int^2_0 \ sin (y^2) dy dx = \int \int _o \ sin (y^2) \ dA \\ \\ = \int^2_o \int^y_o \ sin (y^2) \ dx \ dy \\ \\ = \int^2_o \Big [x sin (y^2) \Big] ^{x=y}_{x=o} \ dy \\ \\= \int^2_0 ( y -0) \ sin (y^2) \ dy \\ \\ = \int^2_0 y \ sin (y^2) \ dy \\ \\ y^2 = U \\ \\ 2y \ dy = du \\ \\ = \dfrac{1}{2} \int ^2 _ 0 \ sin (U) \ du \\ \\ = - \dfrac{1}{2} \Big [cos \ U \Big]^2_o \\ \\ = - \dfrac{1}{2} \Big [cos \ (y^2) \Big]^2_o \\ \\ = - \dfrac{1}{2} cos (4) + \dfrac{1}{2} cos (0) \\ \\$

$= - \dfrac{1}{2} cos (4) + \dfrac{1}{2} (1) \\ \\ = \dfrac{1}{2}\Big [1- cos (4) \Big] \\ \\ = \mathbf{0.82682}$

5 0

3 years ago