ALGEBRAIC FORMULAE

ALGEBRAIC FORMULAE

7th - 11th Grade

20 Qs

quiz-placeholder

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ALGEBRAIC FORMULAE

ALGEBRAIC FORMULAE

Assessment

Quiz

Mathematics

7th - 11th Grade

Hard

CCSS
HSA.APR.C.4

Standards-aligned

Created by

FARIDAH Moe

Used 72+ times

FREE Resource

20 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Given

 2n5=3\frac{2}{\sqrt{n}-5}=3  , express n in terms of p.

 n=(2p+15)2n=\left(2p+15\right)^2  

 n=4p2+225n=4p^2+225  

 n=(2p3+5)2n=\left(\frac{2p}{3}+5\right)^2  

 n=(2p3)2+5n=\left(\frac{2p}{3}\right)^2+5  

2.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Given that

 4m2+3p3=2(52p)\frac{4m^2+3p}{3}=2\left(5-2p\right)  , express p in terms of m.

 p=2+m215p=\frac{2+m^2}{15}  

 p=152m215p=\frac{15-2m^2}{15}  

 p=30+4m215p=\frac{30+4m^2}{15}  

 p=304m215p=\frac{30-4m^2}{15}  

3.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Given

 5p=m(p3)m25p=\frac{m\left(p-3\right)}{m-2}  , express m in terms of p.

 m=10p4p+3m=\frac{10p}{4p+3}  

 m=10p4p3m=\frac{10p}{4p-3}  

 m=10p6p+3m=\frac{10p}{6p+3}  

 m=10p6p3m=\frac{10p}{6p-3}  

4.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Given that

 p5r23=4(3p2)\frac{p-5r^2}{3}=4\left(3p-2\right)  , express r in terms of p.

 r=2435p5r=\sqrt{\frac{24-35p}{5}}  

 r=35p245r=\sqrt{\frac{35p-24}{5}}  

 r=2435p5r=\frac{24-35p}{5}  

 r=35p245r=\frac{35p-24}{5}  

5.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Media Image

The price of a pen is RM x and the price of a shirt is RM y. The total price of items that Amal needs to pay is RM 260. write the equation for the price of the shirts in terms of the price of the pens.

y=2x20y=2x-20

y=202xy=20-2x

y=2x20y=2x-20

y=10+2xy=10+2x

6.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Given that

 q=2(h+k)q=2\left(h+k\right)  , express k in terms of h and q.

 k=2hq2k=\frac{2h-q}{2}  

 k=q2h2k=\frac{q-2h}{2}  

 k=2(q2h)k=2\left(q-2h\right)  

 k=2(2hq)k=2\left(2h-q\right)  

7.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Given

 m=3k2km=\frac{3-k}{2k}  , express k in terms of m.

 k=32m+1k=\frac{3}{2m+1}  

 k=32m1k=\frac{3}{2m-1}  

 k=2m13k=\frac{2m-1}{3}  

 k=2m3mk=\frac{2m}{3-m}  

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