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3 years ago

10

Alex has 175 baseball cards. Rodney has 3 time as many as baseball card as Alex. How many fewer cards does Alex have than Rodney

?

1 answer:

Juli2301 [7.4K]3 years ago

5 0

Times 175×3=525-175=300

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384 = (6 – 2) × 6 - 2 x 2

ioda

Answer:

C

Step-by-step explanation:

If you subtract 6-2 you get 4 then multiply it by 6 you get 24 then multiply 2 and 2 and get 4 and then get 24 subtract it by 4 and get 20 so the answer is c Sorry i cant explain well but at least you have the answer.

8 0

3 years ago

Read 2 more answers

Select the correct answer from each drop-down menu.

vova2212 [387]

As far as I can see, it should be 5/10 is parrots which is simplified into 1/2 .

Hope this helps!

6 0

4 years ago

This is due like right now, please help me!!!!

Varvara68 [4.7K]

$\bold{\huge{\orange{\underline{ Solution }}}}$

$\bold{\underline{ Given \: Rules}}$

<u>The </u><u>sum</u><u> </u><u>of </u><u>the </u><u>number </u><u>in </u><u>each </u><u>of </u><u>the </u><u>four </u><u>rows </u><u>is </u><u>the </u><u>same </u>
<u>The </u><u>sum </u><u>of </u><u>the </u><u>numbers </u><u>in </u><u>each </u><u>of </u><u>the </u><u>three </u><u>columns </u><u>is </u><u>the </u><u>same</u>
<u>The </u><u>sum </u><u>of </u><u>any </u><u>row </u><u>does </u><u>not </u><u>equal </u><u>the </u><u>sum </u><u>of </u><u>any </u><u>column </u>

$\bold{\underline{ Let's \: Begin}}$

<u>According </u><u>to </u><u>the </u><u>Second</u><u> </u><u>rule </u><u>:</u><u>-</u>

$\sf{ 75+b+83=76+80+d=a+81+85+78+c+e }$

$\sf{ 158 + b = 156 + d = 166 + a = 78 + c + e ...(1)}$

<u>According </u><u>to </u><u>the </u><u>first </u><u>rule </u><u>:</u><u>-</u><u> </u>

$\sf{ 75+76+a+78 = b+80+81+c = 83+86+d+e}$

$\sf{ 229 + a = 161 + b + c = 168 + d + e ...(2)}$

<u>From </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u> </u><u>we </u><u>got </u><u>:</u><u>-</u>

$\sf{ 158 + b = 166 + a, 156 + d = 166 + a }$

$\sf{ b = 166 - 158 + a, d = 166 - 156 + a }$

$\sf{ b = 8 + a, d = 10 + a ...(3)}$

<u>Subsitute </u><u>(</u><u>3</u><u>)</u><u> </u><u>in </u><u>(</u><u> </u><u>2</u><u> </u><u>)</u><u> </u><u>:</u><u>-</u>

$\sf{229+a = 161+8+a+c = 168+10+a+e}$

$\sf{ 229+a = 169+a+c = 178+a+e}$

<u>We</u><u> </u><u>can </u><u>write </u><u>it </u><u>as </u><u>:</u><u>-</u>

$\sf{ 229+a = 169+a+c \:or\:229+a = 178+a+e}$

$\sf{ c = 299-169+a-a\:or\:e = 229-178+a-a}$

$\sf{ c = 60 \: and \: e = 51 }$

<u>Subsitute </u><u>the </u><u>value </u><u>of </u><u>c </u><u>and </u><u>e </u><u>in </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u>:</u><u>-</u>

$\sf{ 158 + b = 156 + d = 166 + a = 78 + 60 + 51 }$

$\sf{ 158 + b = 156 + d = 166 + a = 189}$

<u>Now</u><u>, </u>

$\sf{ For \: b, 158 + b = 189 }$

$\sf{ b = 189 - 158 }$

$\sf{ b = 31}$

$\sf{ For \: d , 156 + b = 189 }$

$\sf{ d = 189 - 156 }$

$\sf{ d = 33}$

$\sf{ For \: a, 166 + a = 189 }$

$\sf{ a = 189 - 166 }$

$\sf{ a = 23 }$

Hence, The value of a, b, c, d and e is 23, 31 ,60 ,33 and 51 .

8 0

3 years ago

If 2x + 8 = 0, what justifies 2(x + 4) = 0?

musickatia [10]

Answer:

2(x+4)=0

2x+8=0

8-0=2x

8/2=x

4=x

x=4

7 0

4 years ago

Find the exact values of the remaining five trigonometric functions of theta, suppose theta is an angle in the standard position

nexus9112 [7]

Answer:

$\sin(\theta)=60/61\text{ and } \csc(\theta)=61/60\\\cos(\theta)=11/61\text{ and } \sec(\theta)=61/11\\\tan(\theta)=60/11\text{ and } \cot(\theta)=11/60$

Step-by-step explanation:

So, we know that:

$\tan(\theta)=60/11\text{ and } 90\textdegree$

Since θ is in QI, this means that <em>all</em> of our trig ratios will be positive. Recall All Students Take Calculus. Since it's QI, we refer to A in All. The A tells us that all the ratios will be positive.

Now, let's figure out the remaining ratios knowing that all of them is positive. First, let's find the hypotenuse. Recall that tangent is the ratio of the <em>opposite side to the adjacent side</em>. So, we can use the Pythagorean Theorem to find the hypotenuse. So:

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

Substitute 60 for a and 11 for b:

$60^2+11^2=c^2$

Solve for c. Square both numbers:

$3600+121=c^2$

Add:

$c^2=3721$

Take the square root of both sides:

$c=61$

Therefore, the hypotenuse is 61.

So, our side lengths are: Opposite=60; Adjacent=11; and Hypotenuse=61.

Now that we know the lengths, we can find the other trig ratios:

Sine and Cosecant:

$\sin(\theta)=opp/hyp$

Substitute 60 for Opp and 61 for Hyp:

$\sin(\theta)=60/61$

Cosecant is the reciprocal of sine. So:

$\csc(\theta)=61/60$

Cosine and Secant:

$\cos(\theta)=adj/hyp$

Substitute 11 for Adj and 61 for Hyp:

$\cos(\theta)=11/61$

Secant is the reciprocal of cosine. So:

$\sec(\theta)=61/11$

Tangent and Cotangent:

We are already given that tangent is:

$\tan(\theta)=60/11$

Cotangent is the reciprocal of tangent. So:

$\cot(\theta)=11/60$

4 0

4 years ago