Ask question

Ask question

All categories

3 years ago

10

Jasmine was asked to find the product of 14.8 and 3.65. Her work is shown below. Estimate: 15 × 4 = 60 Actual: 14.8 × 3.65 = 540

.2 Which statements about Jasmine’s work are true? Check all that apply. Jasmine did not make a good estimate. Jasmine did not insert her decimal point correctly when estimating. Jasmine did not insert her decimal point correctly after multiplying the actual amount. Jasmine did not make a mistake. Jasmine’s estimate and actual product don’t match closely enough.

1 answer:

gogolik [260]3 years ago

8 0

Answer:

0 Jasmine did not insert her decimal point correctly after multiplying the actual amount

0 Jasmine’s estimate and actual product don’t match closely enough.

You might be interested in

I NEED HELP ASAP!!<br> like right now plz I will do anything

shepuryov [24]

Answer:

Number 5. = 69

Number 8. = 19

Step-by-step explanation:

3 0

2 years ago

A charity has received donations totaling $2370. However, this is only three-fifths of its goal.

Alinara [238K]

3,950 is your answer. See, you divide 2,370 by 3 to get 1/5 of their goal and then multiply that by 5.

8 0

3 years ago

Read 2 more answers

Either enter an exact answer in terms of pi or use 3.14 for pi and round your final answer to the nearest hundredth.

SVETLANKA909090 [29]

Answer:

18.85 units 3

Step-by-step explanation:

7 0

3 years ago

Drag and drop the correct surface area to match the cube. cube with edge length = 3.2 m 12.8 m²19.2 m²51.2 m²61.44 m²

Ira Lisetskai [31]

- Surface Area of a Cube = 6(L)²
- Surface Area of this cube = 6(3.2)²
= 61.44 m²

3 0

3 years ago

Please help me for the love of God if i fail I have to repeat the class

Elena-2011 [213]

$\theta$ is in quadrant I, so $\cos\theta>0$ .

$x$ is in quadrant II, so $\sin x>0$ .

Recall that for any angle $\alpha$ ,

$\sin^2\alpha+\cos^2\alpha=1$

Then with the conditions determined above, we get

$\cos\theta=\sqrt{1-\left(\dfrac45\right)^2}=\dfrac35$

and

$\sin x=\sqrt{1-\left(-\dfrac5{13}\right)^2}=\dfrac{12}{13}$

Now recall the compound angle formulas:

$\sin(\alpha\pm\beta)=\sin\alpha\cos\beta\pm\cos\alpha\sin\beta$

$\cos(\alpha\pm\beta)=\cos\alpha\cos\beta\mp\sin\alpha\sin\beta$

$\sin2\alpha=2\sin\alpha\cos\alpha$

$\cos2\alpha=\cos^2\alpha-\sin^2\alpha$

as well as the definition of tangent:

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}$

Then

1. $\sin(\theta+x)=\sin\theta\cos x+\cos\theta\sin x=\dfrac{16}{65}$

2. $\cos(\theta-x)=\cos\theta\cos x+\sin\theta\sin x=\dfrac{33}{65}$

3. $\tan(\theta+x)=\dfrac{\sin(\theta+x)}{\cos(\theta+x)}=-\dfrac{16}{63}$

4. $\sin2\theta=2\sin\theta\cos\theta=\dfrac{24}{25}$

5. $\cos2x=\cos^2x-\sin^2x=-\dfrac{119}{169}$

6. $\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=-\dfrac{24}7$

7. A bit more work required here. Recall the half-angle identities:

$\cos^2\dfrac\alpha2=\dfrac{1+\cos\alpha}2$

$\sin^2\dfrac\alpha2=\dfrac{1-\cos\alpha}2$

$\implies\tan^2\dfrac\alpha2=\dfrac{1-\cos\alpha}{1+\cos\alpha}$

Because $x$ is in quadrant II, we know that $\dfrac x2$ is in quadrant I. Specifically, we know $\dfrac\pi2$ , so $\dfrac\pi4$ . In this quadrant, we have $\tan\dfrac x2>0$ , so

$\tan\dfrac x2=\sqrt{\dfrac{1-\cos x}{1+\cos x}}=\dfrac32$

8. $\sin3\theta=\sin(\theta+2\theta)=\dfrac{44}{125}$

6 0

3 years ago