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Anestetic [448]

3 years ago

14

Rodrick and Valentina drove to the coast. Rodrick drove 38 9−− 10miles. Then Valentina drove the last 51 3− 5miles. How far did

they drive to the coast?

2 answers:

Anvisha [2.4K]3 years ago

8 0

15 miles?? im in 5th i don't really know this sorry

ivolga24 [154]3 years ago

3 0

I believe the answer would be 69 1/2 and/or 69.5

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1+4=5 2+5=12 3+6=21 8+11=?

g100num [7]

I think the pattern is that you multiply the first number by the second , then add the first number. SO:
e.g. for the first one 1 x 4 = 4 , 4 + 1 = 5
for the second one 2 x 5 = 10, 10 + 2 = 12...
So for 8+11:
you do 8 x 11 = 88 , 88 + 8 = 96

7 0

3 years ago

I need help with his problem with details.

inysia [295]

Answer:

The solution is y> -1

Step-by-step explanation:

y- 6 > -7

y> -7+6 (add 6 on both sides)

y> -1

The graph should the the first option from the right, with the dot above -1.

Since y is greater than -1, the dot should be on -1 and the arrow should be pointing towards the right. It shouldn't be a coloured dot because the sign is 'greater than' (>) not 'greater or equal to' (≥)

5 0

3 years ago

Dan drew the line of best fit on the scatter plot shown below: A graph is shown with scale along x axis from 0 to 10 at incremen

Anit [1.1K]

0, 3
- 10, 15
= -10, -12
therefore, the slope is 6/5, and the intercept (c) is as supplied, 3.
the equation, y=mx+c or y = a + bx, can be applied here where m or b = 6/5, and a or c = 3.

therefore the equation is y=6/5x+3.

To test this, you can put in y = 10(6/5)+3, which spits out y = 15. This way we know it *should* work.

3 0

3 years ago

Check wether the given value is the solution to the equation<br><br> 3(y+8)=2y-6 y=30

erma4kov [3.2K]

30+8*3=114 2y-6=56.... I think your answer would be 170 but I'm not completely sure

4 0

3 years ago

Prove that sin²∅+cos²∅=1 ?

sasho [114]

Step-by-step explanation:

Let ABC be a right angled triangle where there's a right angle in B . Let the measure of the angle BAC be ∅. So,

$\sin(∅) = \frac{BC}{AC}$

$= > {\sin}^{2}∅ = \frac{ {BC}^{2} }{ {AC}^{2} }$

Also,

$\cos(∅) = \frac{AB}{AC}$

$= > { \cos}^{2}∅ = \frac{ {AB}^{2} }{ {AC}^{2} }$

Now adding the values of sin^2∅& cos^2∅,

${ \sin}^{2} ∅ + { \cos}^{2} ∅ = \frac{ {BC}^{2} }{ {AC}^{2} } + \frac{ {AB}^{2} }{ {AC}^{2} }$

$= > { \sin}^{2} ∅ + { \cos}^{2} ∅ = \frac{ {AB}^{2} + {BC}^{2} }{ {AC}^{2} }$

But we know that

${AC}^{2} = {AB}^{2} + {BC}^{2}$ by applying Pythagorean Theorem

So ,

$= > { \sin}^{2} ∅ + { \cos}^{2} ∅ = \frac{ {AC}^{2} }{ {AC}^{2} } = 1$

7 0

2 years ago